如何在 C 中动态分配字符串数组？

Question

If I have the number of items in a var called "totalstrings" and a var called "string size" that is the string size of each item, how do I dynamically allocate an array called "array?"

This is an array of strings in C, not C++.

Thanks!

Answer 1

NOTE: My examples are not checking for NULL returns from malloc()... you really should do that though; you will crash if you try to use a NULL pointer.

First you have to create an array of char pointers, one for each string (char *):

char **array = malloc(totalstrings * sizeof(char *));

Next you need to allocate space for each string:

int i;
for (i = 0; i < totalstrings; ++i) {
    array[i] = (char *)malloc(stringsize+1);
}

When you're done using the array, you must remember to free() each of the pointers you've allocated. That is, loop through the array calling free() on each of its elements, and finally free(array) as well.

Answer 2

The common idiom for allocating an N by M array of any type T is

T **a = malloc(N * sizeof *a);
if (a)
  for (i = 0; i < N; i++)
    a[i] = malloc(M * sizeof *a[i]);

As of the 1989 standard, you don't need to cast the result of malloc , and in fact doing so is considered bad practice (it can suppress a useful diagnostic if you forget to include stdlib.h or otherwise don't have a prototype for malloc in scope). Earlier versions of C had malloc return char * , so the cast was necessary, but the odds of you having to work with a pre-1989 compiler are pretty remote at this point. C++ does require the cast, but if you're writing C++ you should be using the new operator.

Secondly, note that I'm applying the sizeof operator to the object being allocated; the type of the expression *a is T * , and the type of *a[i] is T (where in your case, T == char ). This way you don't have to worry about keeping the sizeof expression in sync with the type of the object being allocated. IOW, if you decide to use wchar instead of char , you only need to make that change in one place.

Answer 3

char** stringList = (char**)malloc(totalStrings * sizeof(char*));

for( i=0; i<totalStrings; i++ ) {
  stringList[i] = (char*)malloc(stringSize[i]+1);
}

Answer 4

I know this question is old and was already answered. I just want to point out that malloc is a system call and shouldn't be used multiple times.

It would be better to allocate one big chunk of memory and make the array point to it. Something like this :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define STR_SIZE 100 //STR_SIZE doesn't have to be a constant

int main(){
        int i;
        char **arr;
        char *long_string;
        long_string = (char *)malloc(sizeof(char)*STR_SIZE*10);
        arr = malloc(sizeof(char *)*10);

        //Pointing each item of the array to an allocated memory. 
        for (i=0; i<10; i++){
                arr[i] = (long_string + STR_SIZE * i);
        }

        //Initialising the array
        for (i=0; i<10; i++){
                strcpy(arr[i], "This is a line in a\
 paragraph\n");
        }

        //Printing the items of the array
        for (i=0; i<10; i++){
                printf("%s \n", arr[i]);
        }

        //freeing the allocated memory
        free(long_string);
        free(arr);

        return 0;
}

Answer 5

Well, first you might want to allocate space for "array", which would be an array of char * that is "totalstrings" long.

What would then be the starting and final indexes in "array"? You know the first one is 0; what's the last one?

Then, for each entry in "array", you could (if you wanted) allocate one area of memory that is "stringsize+1" (why +1, pray tell me?) long, putting the starting address of that area -- that string -- into the correct member of "array."

That would be a good start, imo.