Allocating (and De-Allocating) Memory for char*[64] in C

I was won开发者_StackOverflow社区dering how I could allocate (and De-Allocate) an array of char pointers (char*[64]) in C. I've looked around on the web for examples but they all focus on other datatypes or one dimension arrays.

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This is answered at The C FAQ.

Here is that code adapted for char* instead of int* with some example data.

#include <stdlib.h>

#define nrows 10
#define ncolumns 64

int main(int argc, char* argv[])
{

    int i;

     //allocate 10 char*
    char **array1 = (char**)calloc(nrows,sizeof(char*));
    for(i = 0; i < nrows; i++)      
    {
        //allocate 64 chars in each row
        array1[i] = (char*)calloc(ncolumns,sizeof(char));  
        sprintf_s(array1[i],numcolumns,"foo%d",i);
    }

    for(i=0;i<nrows;i++)
        printf("%s\n",array1[i]);

     //prints:
     //  foo0
     //  foo1
     //  ..
     //  foo9
     for(i=0;i<nrows;i++)
           free(array1[i]);

     free(array1);

    return 0;
}

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If you want a fixed-size array, chances are pretty fair that you don't need to use dynamic allocation at all. If you are going to allocate it dynamically, you'd do something like:

char **array;   
array = malloc(64 * sizeof(char *));

When it's time to free it, you'd just do like usual:

free(array);

Note, however, that if the pointers in the array point at anything (especially dynamically allocated memory) you need to deal separately with freeing that memory -- usually before you free the array of pointers.

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You allocate memory on the heap with malloc:

char** pPointers;
pPointers = malloc(sizeof(char*) * 64);

Now you have an array of 64 pointers to character(s).

You deallocate memory with this call:

free(pPointers);

You will probably assign values to your array:

pPointers[0] = "abc";
pPointers[1] = &cSampleCharVariable;
pPointers[2] = strdup("123");

Freeing this memory depends on what the pointers point to. In the samples above only pPointers[2] can be deallocated.

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If you have the type given in your question:

char *foo[64];

Then you don't need to explicitly allocate memory for foo - it's a normal array. To allocate memory for the pointers in foo, just use a loop:

int i;

for (i = 0; i < 64; i++)
{
    foo[i] = malloc(somesize);
}

(You shouldn't really use the magic value 64 in the loop, though; either use a #define FOOSIZE 64, or use sizeof foo / sizeof foo[0] to determine the size of the array).

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At the risk of being obvious (and less general than the other answers), sizeof ( char * [64] ) equals 64 * sizeof ( char * ), so you can just write p = malloc ( sizeof ( char * [64] ) ) to allocate the memory and free ( p ) to free it.

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Doing the 65 allocations of the accepted answer will work, but If you know the lengths of the char data ahead of time you can make it a hell of a lot faster by doing something like this.

Assume cItems is an int that is the number of elements in your array, say 64. Assume aItemSizes[] contains the lengths of the char data for each element in your array:

int iItem;
char **aItems, *pData; 

for (cbCharData=iItem=0; iItem<cItems ; iItem++)  
  cbCharData += aItemSizes[iItem];
aItems = (char**)malloc(sizeof(char*)*cItems + cbCharData); 
for (iItem=0, pData=&aItems[cItems] ; iItem<cItems ; pData+=aItemSizes[iItem], iItem++)  
  aItem[iItem] = pData;

Now aItem[iItem] points to a memory location that can handle up to aItemSizes[iItem] chars and you only needed one allocation, which is WAAAAAAAAAAAAAAAAAAAAAAAAAAY faster. I apologize for compile errors cuz i didnt test this. To free it you only free aItems.