C malloc array of strings. Dynamically allocated string arrays in C.

C malloc array of strings When given a single character buffer, fgets() can only store the null terminator. C There is no way to know the size of a dynamically allocated array, you have to save it somewhere else. Then you read the second string into that space, and make w_arr[1] point to that space. Also, please do not cast the return value of malloc() and family. Use strcpy() instead. It has the following signature: char *fgets(char *restrict s, int n, FILE *restrict stream); As it wants a stream, you just should give stdin. Invalid Operations in If you need to extend an array piecemeal instead of allocating it all at once, you would use realloc() instead of malloc(). Arrays are zero-indexed: var[1] points to the second element in the array, not the first. – Daniel Kamil Kozar Commented Dec 6, 2015 at 21:09 The strings will be stored in an array of 6 char pointers. Memory Allocation in c with string. That's why when you write a function that takes a pointer where you pass an array, the function should also take the size of the array, otherwise the size cannot be calculated, for example: Yes, if you need to do this without memmove, you can do it with a simple loop. This is the most flexible method, but it makes the caller responsible for free()ing the allocated I am trying to take multiple string input in an array of char pointer,the no. This in-depth guide will teach you how to leverage malloc() to Here is a practical C program that demonstrates dynamically creating an array with malloc (), resizing it with realloc (), populating it with data, then freeing the memory: int n = 5; // In this comprehensive guide, you‘ll learn how to create dynamic arrays in C whose size can grow or shrink as needed. If there is a If you ever find yourself attempting to use sizeof() for anything that is sized dynamically, you're doing it wrong. C Using malloc on a char string. The strings used for the types of ants are String-Literals and on all but non-conforming compilers are read-only (non-mutable). Hot Network Questions How rigorous would sterilization have to be for a Europa Lander? Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Visit the blog I read the previous questions on dynamic arrays in C however I was not able to relate the answers to my question. There are several ways to check for memory leaks in a program. You need to find the length of the string for the malloc. struct line* array = (struct line*)malloc( First, you need to allocate array of pointers like char **c = malloc( N * sizeof because c 2-d arrays are just and array of arrays. Using malloc() to create dynamic arrays removes these size constraints and leads to more adaptive programs. strcat takes care of the null bytes for you. That's what I have Arrays are a vital data structure in C. Using malloc for an array of strings. Make duplicate of all words in string. Referencing string array in C struct Hot Network Questions Is it possible to generate power with an induction motor, at lower than normal RPMs, via capacitor bank or other means? The main difference between {} and malloc/calloc is that {} arrays are statically allocated (don't need freeing) and automatically initialized for you, whereas malloc/calloc arrays must be freed explicitly and you have to initialize them explicitly. You need to allocate an array of char * as well as an array of char for each component word that is stored in the array of char *. Under MacOS, we can compile the program with the -fsanitize=address option. str2=malloc(strlen(str) + 1); // +1 for the null character to terminate the string Share. Lastly: C can't assign strings using =, you'll have to use strncpy or strncat, which requires the string. dynamic memory allocation for strings. You may use any combinations of the showed initializations for any element of the array. Trying to alter them will invoke undefined behaviour. e. I'm trying to create an array of structs (of arrays) and am a bit unsure of the malloc required. type *p = malloc( sizeof(*p) * len ) ); for ordinary buffers. This line: char **array = malloc(10); allocates 10 bytes, however, remember that a pointer is not the same size as a byte. malloc and family allocate memory in chunks of bytes. Currently, I have some confusion in realloc an array string. Hence, you need to specify that same number again in the call to secure Write a function that will append the null terminator to the end of a string. h> /* realloc free */ #include <stdio. The allocation strategy in the function is wrong, too, therefore. The char* your storing in each node points to memory in the stack variable string[50], which is overwritten with each new fgets() call. char (*lines)[1000]; However your structure is in fact declaring an array of 128 char pointers, which is conflicting with your malloc statement. . As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member. This was really tough to explain so here is an example: Using fgets and malloc, allow the user to enter multiple lines of strings. inputting strings into a multidimensional character array? 0. Besides, gets() is unsafe, as there is no size limit. h> /* assert */ #include <stdint. Made the changes, had to add (**char) on the realloc. Commented Dec 7, 2015 at 20:03. Still it is NOT the same. malloc-ing a char array. Here are some key takeaways: malloc() allocates memory for arrays on the heap ; Array size can be changed by reallocating You allocated the char *[] array, now you need to allocate each of the strings that the user input, ie each words[i] needs to be allocated. a="abc"; This assigns a pointer to a constant string to your char* a, by doing so you loose the pointer to the memory allocated in the first line, you should never free constant strings. Then you read the third string into that space, and make w_arr[2] point to that space. The most efficient solution is to compute the size of the concatenated string, and then strcpy the components into the correct place. Before C99 introduced VLAs, you also needed it to perform allocation of a dynamically-sized Something like char ** is not a 2D array and cannot represent one. Edit - sizeof() sizeof() returns the size of a variable. So the parameters str1 and str2 are actually pointers to pointers to char. Therefore you need to make sure you allocate an array of sufficient size by using the size of the related type: You have multiple problems. What you get from malloc is a block of raw memory, which does not contain a properly constructed object. For example if n is 2 you'll have Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company An array of strings allows you to store and manipulate text in C. That's the type of the argument to malloc. '; array[4]='9'; And you'd like to have this: That's because you are the same pointer to char for all positions of the array. Each element in your 2D array will then point to the string, rather than holding the string itself. s1, s2, . Here's an example that creates and extends an array one element at a Thanks for the prompt reply. pointer to char[100] array to your function. sizeof() is used repeatedly in this code incorrectly. Allocation memory for array of strings. String. The trick to this is having the loop move each element one forward, starting from the last one. You can also just point to string literals, but remember that literals may not be writable; i. Many object oriented languages (eg. Any heap-allocated memory must be freed. Since strings in C consist of one byte per character and ideally have to be null terminated you need to make sure you have enough memory to hold that. The inventory should be "any-sized" said my professor. The array should be declared as: char *str_array[6]; This type of array only stores pointers to a string, and nor the strings themselves. Declaring the parameter as type char [] will also work for both statically and dynamically allocated argument values, but char * is more commonly used for specifying the The better method should be a 2d array of pointers, so I can dynamically allocate memory for every new string. ). Neither. Improve this answer. For home-work, I need to define a function that allocate memory to an array of strings (which is into a struct). Code: When you use malloc() the memory is dynamically allocated at run time, so you need not fix array size at compile time also u can make it grow or shrink using realloc() None of these things can be done when you do: char some_memory[] = "Hello"; Here even though you can change the content of the array, its size is fixed. When you do: words[i] = txt; You are assigning a pointer. This works in the case of an array because the array name is converted to a pointer to its first element. You can of course still store your n x n elements in it:. I am trying to swap 2 elements in an array of pointers, and these pointers point to strings of different length. There are some issues with your code. You don't assign strings after malloc(). if you names are no longer than 25 characters, then you could safely declare name[26]. You should stick with pointer to char array to use it as array of string. Creating an array of strings using malloc in C. We‘ll be using the malloc () function to allocate array memory on the You cannot store an array of strings in C, as a string is a variable-length datastructure, not a simple type. The assignment doesn't copy anything, just makes test point to that area of memory. Let's look at what you need to do to read and store an unknown number of strings (or any object for that matter). 2. g. @PCR It's not really "the address of a string" - it's the address of some space in memory. In C there are pointers to a string. It all seems to work but the program behaves strangely. Unlike static allocation, dynamic memory allocation happens A string array in C can be used either with char** or with char*[]. There are three potentially offending pieces of code and I can't figure out where I'm going wrong. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types You can't tell: it's your job to keep track of the size, or to use a magic value (NULL, say) to mark the final element. In this tutorial, you'll learn to dynamically allocate memory in your C program using standard library functions: malloc(), calloc(), free() and realloc() with the help of examples. Ultimately, you are copying only the null terminator. If you want to store an array of several strings, you need to malloc() an array of char * PLUS the several strings. Also, you are using strcpy a lot of times. C99 has a special construct for such things, called flexible array member of a struct:. quick and dirty example :) (Should realy init the elements) There seem to be some confusion about what a string is in C. (Well, actually there are none, but there are three common ways to return a pointer to a string, which the caller can then use to access the string. Each index of the array will point to each line of the text that is entered from the user. Ref: As per C11 standard, The number of characters do not come close to the size of the array, therefore I need to use malloc() in order to get the exact array size. Comments are in lined for understanding, please get back in case of any clarification needed. However, you cannot return values stored on the stack, as in your function. storage space for your string). Whereas converting the statement char *p = malloc( len + 1 ); would require more thought. Go through the below program which explains how to allocate memory for char** and also how to free the same. And I would like a malloc'ed array of malloc'ed strings char *myarray[] such that: myarray[0]=="valgrind" myarray[1]=="--leak-check=yes" Edit I have to assume that there are an arbitrary number of tokens in the inputString so I can't just limit it to 10 or something. The strlen() function returns the length of a string, which is the number of characters preceding the terminator, therefore if you want to calculate the Yes, you have to free() every block you obtained from malloc(). Here is a toy example program that asks the user to enter some strings. The code compiles and runs the problem now is that after the loop if I print the value of lnames[0] and lnames[1] they have the same value, the last file in the folder (hence last string in the loop), so it appears to be overwriting the entire array with each entry. To return an array of (char pointer) (or (char pointer pointer) ) you need to either malloc a new string and store the address of that in env_vars, or have env_vars store the address of args[i]. matrix[i * n + j] = rand() % 10; If you prefer, you can set up a 2D structure by following the advice given in How do I work with dynamic multi-dimensional arrays in C?. b) Why are you seeing that each array element occupies a different amount of bytes. The problem is that you are using sizeof when deciding how many characters to copy. The length of each string is given: MAX_WORD_LEN+1 (=10+1) I have to allocate memory for len number of strings, len is recieved in the input. * The string memory is allocated with malloc(), * so the caller must release it using free(). I just want to call the build_array() function and return the array of strings. Since that's all that's giving it structure, you can wipe out the structure by There are strings in C, though not a string type. If I have this: char** str = (char**)malloc(100*sizeof(char*)); str[0] = (char*)malloc(sizeof(char)*7); //allocate a space for string size 7 //some other code that make the array full My question is, if I want to realloc str[0] to size 8, do I need to realloc both str and str[0] like malloc a char**, and then malloc the space needed for each of the strings, storing the pointer to them in respective indices of the array created by the first malloc. About the mistakes First - calloc() has TWO parameters and not one, like malloc(), as in the following signature: Strings are fundamental to almost every C program. Viewed 3k times 0 . Syntax of Array of Strings. I want to create an array of strings called arguments that copies entries from an array of strings called words (from words[1] until the end). by you having a variable strings_populated or something that tells you you've put strings in some N of the potential locations. This in-depth guide will teach you how to leverage malloc() to create, manipulate, [] I'm a noob so don't be hard on be. You have several ways of creating C Malloc array of characters. The name "malloc" stands for To have a truly unlimited buffer in C, (or limited by the amount of memory and a size_t,) you could build up the memory allocation incrementally. So if you concatenate str1 to str2, strcat(str2, str1) will locate the terminating null of str2 and start copying str1 at that location until it finds the terminating null of str1, and copies that, too, and stops. 3. I can guarantee that the strings fit within MAX_STRING_LENGTH characters, but I don't know at compile time how many strings will be in the For the actual strings, (that is: character arrays) you use malloc() to reserve memory. A I was not really capable to understand already available solutions as it had a slightly different way to use it than I want mine. You can add as many strings of arbitrary length as you want, until your computer runs out of memory, at which Point 1. Instead of malloc-ing object, use new. * On memory allocation error, NULL is returned. However, you allocated a fixed number of characters which is not known to sizeof operator: sizeof StrPtrArr[0] is equal to the size of char pointer on your system (four bytes, judging from the output), not 10 + 1. 4. So if you wrote str = "foo", you'd be trying to assign the address of the first character It looks like you are wanting to allocate for an unknown number of strings. 66% off Learn to code solving problems and writing code with our hands-on C Programming course. I just want to point out that malloc is a system call and shouldn't be used multiple times. But of course, malloc/calloc arrays don't go out of scope and you can (sometimes) realloc() them. You're saying you have this: char array[20]; char string[100]; array[0]='1'; array[1]='7'; array[2]='8'; array[3]='. The first . h> int main (void){ int i, n, l; char **p; To tackle your second questions, by executing the statement copy++ you have changed the value of copy (that is, the address in memory that holds a char array) so that by the time you print it out, it is pointing at the end of the array rather than the beginning (the value returned by malloc()). , a 3 MB local stack array is a bad idea). Using array of string in C. Using the malloc() memory allocation function provides more flexibility and control when working with strings in C. type *p = malloc( sizeof(*p) * ( len + 1 ) ); for zero-terminated strings and. h> /* C99 SIZE_MAX */ #include <stdbool. Memory allocation for char pointer in a struct. You pretty much can't return an It's a 1D array, so you have to treat it as a 1D array, and not as 2D. You must be intend to use the variable 'lines' as an index for the allocated memory, since the allocated memory is actually an array of 10*30=300 chracters. Note that the c runtime probably keeps track of the amount of memory allocated, but this is not exposed to you in any portable way. You cannot change the contents There is no way to do this in C. Ask Question Asked 1 year, 10 months ago. Use strcpy(a,"abc"); instead of a="abc"; to move the string into your allocated memory. Dynamically allocated string arrays in C. It doesn’t Initialize memory at execution time so that it has initialized each block with the default garbage value To create a dynamic array of strings in C, we can use the concept of double pointer and dynamic memory allocation. Viewed 4k times 5 I'm currently learning about strings, pointers and arrays in C. The double pointer is the pointer that stores the memory To create an array of strings using the “malloc()” C standard function, first create a simple C program and declare two arrays, one of which is a pointer array. h> int main() { // Creating array of stri I guess you want to allocate 10 arrarys, whose length is 30 characters long for each, with only one malloc() function call. If There are three common ways to return a string from a function. If you really want to read the word into a buffer (like txt) and then put it into the array of strings, you need to copy the contents of the buffer string to the string in the array, like so: I will read in two set of char* (or strings) using strtok, and since those two set of chars are related, (address : command\\n) I decided to use a structure. The user will signal the end of the strings with a '. It returns a pointer of type void which can be cast into a pointer of any form. In this case, you should have enough memory to store the original string contents and the "NUL" character (length+1). You must allocate space for each string using malloc(). It doesn't matter if the variable is an int, array or 2d array. I need to malloc the string so the symbols and gibberish won't be included. You cannot free it. Malloc a string array - C. I tried to initialize a double pointer and use a function to insert elements, it allocates space for another column(new string) and set the length Use it like arr[index], just as if it were declared as an array. Use the size_t type for storing array indices and object sizes. 41. 0. One way to achieve this would be to allocate a 3d array of char as: char x[50][7][MAX_LENGTH]; You can think as having a matrix of array start (of pointers) and then, another dimension to give depth to your matrix (i. h> /* strcpy */ #include <assert. What is the best way to initialize the string array in c? 1. The tokenizing part is working fine but the dynamic memory allocation seems not to be, based on a segfault when the print loop is uncommented. sizeof looks at the type, but you can't obtain a complete array type (array type with a specified size, like the type int[5]) from the result of malloc in any way, and sizeof argument can't be applied to an incomplete type, like int[]. * The input string array must have NULL as last element. It is usually represented by char*. More like a missing-malloc problem. For strings, you always need at minimum the number of chars to store + 1 (for the When you are doing string_copy(my_array), you are passing a char (*)[100], i. In C, the notation x[y] is exactly equivalent to *(x + y). Keep that in mind. For my homework, I would have to make an array of strings dynamically using malloc. It's just the only way to use the syntax a[i][j], while still allowing both dimensions to be unknown at compile time. I want to malloc this array of struct, since I do not know the exact number of file before strating the program. ' on a newline. Trouble duplicating c string and storing in new character array. Memory is allocated for the strings in the get_string() function, then pointers to the strings are added to an array in the add_string() function, which also allocates memory for array storage. of strings is also taken from user. This is a quite common misconception because an array automatically decays as a pointer when used as a function argument and it is implicitly convertible to a pointer. Note that by doing this you will still run into trouble using the C library of functions for manipulating strings. An array is a continuous memory block of elements of the same type with a linear addressing scheme. Modified 12 years, 6 months ago. Try changing the line with malloc into: int n, i; scanf("%d", &n); struct subject *ptr = (struct subject*) malloc(n*sizeof(struct subject)); You also have to change your input into: There are some topics that partially overlap with this but I am still seeking an answer. I pass the pointer to the array into this function, first freeing the individual strings, then the array itself. Share You cannot copy the input C-strings to the output, unless you want to return a (char pointer) that points to all the strings concatenated together. @Boninissimo, it's a bit more crypric but it's +- easy to understand, arr is a pointer to an array of 40 chars as such sizeof *arr (the dereferenced pointer ) is 40 bytes in size, this multiplied by n will give you the block of memory you need, now as arr is a pointer to 40 chars, incrementing it will make it point to the next block of 40 chars. Pointers, Arrays, Strings and Malloc in C. malloc will not care if you allocated those 1000 bytes to hold a string or any other data type. I'm working on an assignment and I have to write a function that takes in a dynamically allocated array of strings and an index, and removes the element at that index. h> /* stdin fgets printf */ #include <string. , sn};. These functions are defined in the <stdlib. return toReturn } This will obviously return warnings and will not work properly since the memory location is going to be free-d after the function is finished. For a 2D array each element is an array itself. So, decide what you want: An array of fixed-length buffers To allocate memory dynamically, library functions are malloc(), calloc(), realloc() and free() are used. Hot Network Questions Burned washing machine plug I am trying to allocate memory for an array of C strings. You are thinking correctly that you will want to use a pointer-to-pointer-to char (e. array of char pointers, because you have defined your function that way. Let's take a look at an example: [GFGTABS] C #include <stdio. This article gives good detail. h> void print_arr(char* arr[]); void clear_arr(char* arr[], int size); int main() { int number important: I don't have a deep understanding of malloc so try to put things as simple as possible Greetings, I want to create a dynamic array of strings (using malloc) and then print those strings out (for starters). You need to allocate more space per string, or read the line into a big string and then duplicate (strdup() instead of the original malloc() is easiest; you could use realloc() since you already did malloc() for each string) the line into the stored string. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Your function return type is all wrong; it needs to be char **, not just char *. But static arrays with fixed capacity can be limiting. removing the newline character and then want to store each command delimited by a space in a dynamically allocated array of strings. 5. That requires keeping track of the sizes of the strings, which is a bit of overhead, or scanning the length of each component string twice. Struct with the array of strings definition (given): I want to make an array of string using malloc, and then clear ALL the allocated memory, I believe that I use malloc correct but I can't understand what I'm doing wrong when I try to clear it: #include <stdlib. 5 String literals. There are two possible interpretations of your question: To fill a destination char array with the concatenation, or; To (m)allocate a new (big enough) string and fill it with the concatenation Some things to keep in mind: If you are going to use sizeof in malloc calls, do yourself a favour, and use var = malloc(40 * sizeof *var);. h> /* C99 bool */ /* Every time I allocate the memory for a 2D array first I create an array of int** and then with a for I allocate the memory for each element. Viewed 101 times 1 I understand that assigning memory allocation for string requires n+1 due to the NULL character. Malloc not allocating space. 1. The end result is What the compar function gets are pointers to the elements in your array, which in this case, are pointers to char. h> #include <string. Then you read the first string into that space, and make w_arr[0] point to that space. Strings in C need to be NUL-terminated, that is, there is always a \0 character after the last character of the string. example *ex = new example; Your original code can be forced to work with malloc as well, by I am trying to make a function that takes a string and a pointer to an array of strings and malloc() the array of char arrays and copies each individual word of the string. The first step is to allocate a single array, then loop through it allocating arrays for each column as you go. First I define my struct, typedef struct { char *str1, *str2, *str3, *str4; } player; Then in main I need to initialize the structure, and malloc the strings inside of it, player1 player; player1. I however am having a lot of trouble with the correct way to allocated and reallocate memory Declaring Static Character Arrays (strings) When you know (or have a reasonable idea how large your array needs to be, you can simply declare an array of sufficient size to handle your input (i. (with a twist to separate each string into individual tokens before saving to the array) It is worth understanding this process in detail as it will serve as the basis for just about any any other circumstance where you are reading an unknown number of Create an array of string pointers. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Visit the blog @AndrewS: The complete answer won't fit into a comment, but basically it's an artifact of how C treats array expressions; under most circumstances, an expression of type T [N] is converted to an expression of type T *, and the value of the expression is the address of the first element. Point 2 [Without malloc()] string literals (here, "Test") are usually stored in read only memory. Dynamic size counts must be managed by you; the allocation size requirements in bytes can be managed using those counts in conjunction with a sizeof() of the fundamental After malloc, the contents of memory are undefined, so your current code could do any number of things, most of them incorrect. By the way, you probably don't want to be calling srand() every time you create a You can't malloc a class with non-trivial constructor in C++. And as @Nyan suggests in a comment, you could also do. For what it is worth, there are much easier ways to do this than you're I am stuck in how to fill a pointer array with strings using malloc. C - how to malloc struct containing string and array? 3. There is no standard library function to do that. char **arr;), but the "train fell off the tracks" so to speak when you went to implement the logic. Modified 13 years, 7 months ago. Golden rule of C programming. Instead of something like this; char string[NUM OF STRINGS][NUM OF LETTERS]; Is it possible to dynamically allocate how many strings will be in the array with malloc just like when you dynamically allocate memory for char pointer? You can use strlen for the string length. A string literal has the type array of characters with the length set to contain the characters and the '\0' (nul-terminating characer). So use Allocate a block of memory through a call to malloc() Keep track of the size of input; When ever you need a increament in buffer size call realloc(ptr,size) Share. You asked for 11 strings so char Answers[10][100]; is incorrect and you should type char Answers[11][100]; instead, that was the reason why it skipped input. char *a=malloc(sizeof(char)*4); You allocate memory so you should free it. But string handling can be tricky in C due to null-terminated character arrays, pointers, and memory management. The function strcpy does this work for you. name = (char **) malloc (sizeof (char *) * 128); using the structure below. Only you know that there's a tree-like structure that could be freed recursively, that knowledge is not anywhere in the C runtime heap, so the heap manager has no idea about without much thought. I tried to write a program where an array holds three pointers to string addresses. For example: int ** arr = malloc(N*sizeof(int *)); for C malloc() method. , sn: Strings to be stored. , you may not be able to do something like: strs[j] = "foo"; strs[j][0] = 'b'; // may not be allowed on string literal Are you looking for a jagged array (array of arrays of different length), or X*Y fixed size array? And are you looking for 2-dimensional array of strings(==pointers to char), or are you looking for array of pointers to arrays of strings(==pointers to char), or are you looking for 3-dimensional array of char with no pointers in the array? – So, you need a 2d array of strings (char arrays). Initializing an array of strings dynamically in C. The source string will always have only a single space as the delimiter. h> header file. h> int main() { // Creating array of stri In C, an array of strings is a 2D array where each row contains a sequence of characters terminated by a '\0' NULL character (strings). You've declared an array of char *s but what you really want is a pointer to an array of chars. If you use the C string library, it just does the right thing with strings that terminate in null. It's all about reducing mental overhead. When I print the array without using the malloc function is prints symbols and other gibberish along with the string. So something like "apple" has a length of 5, but contains 6 characters total, including the terminator. Use the line. However, the question is what if you allocate 10 chars but enter an 11 char string? Trying to give an answer here, there's probably one thing to clarify first: an array is not a pointer. I'm having trouble with malloc and don't really understand how much I should malloc. A C string is an array of bytes that starts at a given address in memory and ends with 0. In order to use it as a pointer to your structure you have to cast it first. You should The “malloc”or “memory allocation”method in C is used to dynamically allocate a single large block of memory with the specified size. How to malloc an array of string in ANSI C? Ask Question Asked 12 years, 9 months ago. No. C program malloc with array of strings. In C++ I can allocate an array of strings It is not so clear how your struct type is declared. You have to either pass a parameter with size information, so that malloc() and free() can be called in the called function, or the calling function has to call free after malloc(). #include <stdlib. strcpy copies each character up to an expected null terminator into your new buffer. This is what I have so f C malloc array of structs of strings. How to you know the size big mystery! Eith you allocate a unique big string and ensure the user doesn't input more than that, input in it, allocate words[i] with the now known size, and copy the big string into this new allocated area. In case of "Zorro" it will occupy 6 bytes: 5 for the string, the 6th for the 0. String literals are usually stored in constant/read-only sections of memory. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company In case you arrive in this page, wanting to create an array like int myarray[n][M] (which is slighly different from the question since they want an array of string), where M is fixed and n can vary (for example if you want an array of coordinates), then you can just do: Your array_of_strings is nothing but 10000 contiguous bytes. A moment's reflection should tell you why this is necessary. @Chris: that said, it's guaranteed to be big enough for 5. In C: "a string is a contiguous sequence of characters terminated by and including the first null character". The standard section for string literals is C11 Standard - 6. When I run Terminology quibble: this is not what C usually calls a "multidimensional array". Modified 1 year, 10 months ago. h> #include <stdlib. The other kind of multidimensional array an array of arrays, instead of this array of pointers to (the first elements of) arrays. Here is my code: struct get_data{ int sequence; int mask_ID; char *name; float intensity; float angle_correction; double points[10000]; float X_interval; }; struct get_data all_data[number_of_file]; You use malloc when you need to allocate objects that must exist beyond the lifetime of execution of the current block (where a copy-on-return would be expensive as well), or if you need to allocate memory greater than the size of that stack (i. I am trying to allocate an array of strings (using malloc). Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company This way you can allocate as much or as little memory as each string in the array needs, and you can resize your strings if necessary. The problem is that I'm new on c and I have a lot of confusion about that technique. Another function handled allocating memory to the array and the strings, the swap function will simple take an char** array and swap the elements I need to swap. r: Maximum number of strings to be stored in the array m: Maximum number of character values that can be stored in each string. Allocate space for the string using malloc() inside the function. Otherwise, you'll be leaking memory. Dynamic array of pointers. Put this string into array at the correct offset. "A pointer to a string is a pointer to I'm running into buffer overflows when reading a file in C and copying character arrays. 4. The first is that you in main you are declaring dictionary to be an array or arrays which means it's already allocated (the compiler allocates memory for the arrays for you), meaning your assignment is wrong. Hot Network Questions is wrong, because ptr is pointer, not an array, so sizeof ptr returns you the size of a pointer, not the total amount of bytes needed by arr. * Returns the pointer to the resulting string. – Jonathan Leffler This is the classic question of how do I handle dynamic allocation and reallocation to store an unknown number of strings. It looks like you want an array strings, each string holding at most 1000 characters. The “malloc In C, an array of strings is a 2D array where each row contains a sequence of characters terminated by a '\0' NULL character (strings). First attempt: char* loadValues (char* str) { char* toReturn[5]; . some operations here . So if you do malloc(1000) you get 1000 bytes. The key benefits are no size Using the malloc() memory allocation function provides more flexibility and control when working with strings in C. You do it by traversing the array of pointers and caling free() on each element and only then freeing the array itself. But your function is expecting a char *[], i. When you have a value like "Zorro", it's a C string. After that, utilize the “malloc()” function by using the “pointer-array = (cast-type*) This guide covers the step-by-step process for using malloc() to allocate memory dynamically for creating array of strings in C programming. You're only allocating space for the node; not the char data you're sucking out of your string[50] buffer with strtok(). If you do concat[0] = 0 before the strcat 's, then your code works but will have to search for the length of str1 three times -- once for strlen , again for the first strcat , and last for the second strcat . int *arr = malloc (MBs * 1024 * 1024 / sizeof(int)); Also, you suggest to malloc() for the name member, the name member is an array and you cannot assign to it, it has enough space for a 31 character string which you can use by assigning to each element, or using strcpy()/memcpy() to copy a complete array to it. Of course, to free an array of strings, we must free each individual allocated string before freeing the array itself. You must cast them like this: int sortstring( const void *str1, const void *str2 ) { const char *rec1 = *(char**)str1; const char *rec2 = *(char**)str2; int val = strcmp(rec1, rec2); return val; } I need to allocate arrays of chars by malloc() and then print them. MY PROBLEM: So, in short: I am making a text-based adventure game. #include <stdio. str1 = malloc(100); // and the rest Both statically declared and dynamically allocated arrays of characters can be passed to a char * parameter because the name of either type of variable evaluates to the base address of the array in memory. Strings with malloc in C. For that, your declaration should be. It is used to store multiple strings in a single array. Here is some examples of how character arrays can be initialized in C. Allocating memory for a string in C using malloc. In C++ I can allocate an array of strings (say 10 strings) very easily as: string* arrayOfString = new string[10]; however I don't know how to do it in ANSI C. You can fix this by making changes so that your function (string_copy()) expects a char (*)[100], instead of a char *[]. Okay, lots of misunderstandings, I guess. You also cannot modify that string. If you want the malloced memory to represent a C string, set first element to 0, a->array[0] = '\0'; – DBug. char arr_name [r][m] = {s1, s2, . Most pieces of code there are incorrect. So every single word[i] is the same string (txt). If you want to return the string array, you have to reserve it dynamically: Your malloc and free stuff is fine, but you are orphaning the original string when you try to do assignment like this: nameslist[0]="John"; Instead you should use strcpy : In c you can not allocate an array of string directly. You have a "pointer to pointer". C++) handle memory in such a way as to do what you want to, but not C. Malloc returns a void pointer type. Array of array of string in c with malloc. – Step-by-step example to create dynamic string array with malloc() Sample operations like sort, search on string array; Best practices for optimal usage and performance ; Buckle up for an in-depth tour of malloc() in context of string arrays in C! Overview of Dynamic Memory Allocation. Its data only has any organization by virtue of however your code understands it to have organization, e. It would be better to allocate one big chunk of memory and make the array point to it. For character strings, the standard library uses the convention that strings are null-terminated: a string of n characters is represented as an array of n + 1 elements, the last of which is a "NUL" character. Note that you might also need to use realloc first, to expand the size of the allocated array so that it can fit the new element. If this is actually your intent then your malloc call should be like so; thisOne. In debug i see that when i fill the 1st pointer of array with a string, when its about to go to next pointer in the array it pass the next string in both 1st and second element seems like when i use ptr[i]=buff; the ptr keeps showing in For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Those functions will only test this string: "how" since the null terminator would be at the end of how. For example, see the following program Because you have initialized pointer of type char** with 4*sizeof(*members) I think you are planning to store 4 strings, so recommendation is to allocate memory for all 4 strings, like: char** members=malloc(4*sizeof(*members)); int i; for(i = 0; i < 4; i++) members[i] = malloc(100 + 1); // 100 is a max length of string // actually length of The first version doesn't create a string on the stack, but you're correct that you're not allowed to free it after the assignment. Another problem is that the array is probably to large, since most compilers on most systems allocate local variables Malloc a string array - C. Initialize array of strings. Ask Question Asked 13 years, 7 months ago. Using a sample main function In your case, as it is a 2D array of strings, it could be declared as char * array[ROWS][COLS], and then you could asign a string to a specific element this way: array[nrow][ncol]="Your String". typedef struct { char ** name; } arr; Hope this helps. And as a string terminated with a null character, the array of pointers is terminated by a null pointer: And as a string terminated with a null character, the array of The Backstory: I created a function to destroy an array of strings in c. h> #include <stdio. I have written following code but it does not work properly, Please if somebody could Malloc a string array - C. * If the input string array pointer is NULL, * NULL is returned. Edit no, that is not memory copy, but pointer assignment. Here, arr_name: Name of the variable. h header In C, an array of strings is a 2D array where each row contains a sequence of characters terminated by a '\0' NULL character (strings). Any attempts to use that memory as a "real" object will fail. In C, a null terminated sequence of chars is considered a string. Instead, use fgets(). I know this question is old and was already answered. purde eae tqgry fknve qsymzy afggknz uujyhgm eko tjcud knzvu