Pointer to a Pointer to a Structure and malloc memory for array

Question

I'm creating a pointer to a pointer to a structure to create a dynamic array with malloc in C, but I get a segmentation fault calling the struct array. Here is a quick rundown of my code:

#include <stdio.h>

typedef struct {
    int test1;
    int test2;
    }testStruct;

int main() {
    testStruct **neato;

    neato = (testStruct **) malloc( sizeof(testStruct *) * 5);
    // Array of 5 for convience

    // any neato[x]->testy call results in segmentation fault.
    scanf("%d", &neato[0]->test1);    // Segmentation fault

    return 0;
    }

I tried other calls like (*neato)[0].test1 and all result in segmentation fault. This is obviously not the proper way to do this or my GNU compiler is seriously outdated.

Do not cast the return value of `malloc`. See: http://c-faq.com/malloc/mallocnocast.html — Matt Eckert, Apr 05 '12 at 02:44

Ed S. · Accepted Answer · 2012-04-05T04:39:42.740

5

You've allocated enough memory for 5 pointers. You have not however initialized the pointers, so they are garbage. Allocate the pointers and then proceed to initialize each pointer.

int elems = 5;

neato = malloc(sizeof(testStruct *) * elems);
for( i = 0; i < elems; ++i ) {
    neato[i] = malloc(sizeof(testStruct));
}

On a side note, I don't see a need for an array of pointers here. Why not simply allocate enough space for 5 testStructs (i.e., neato becomes a testStruct*) and pass the address of that pointer to the function that initializes it?

edited Apr 05 '12 at 04:39

answered Apr 05 '12 at 02:40

Ed S.

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D'oh! Went straight over my head. Thanks for your help. – T.Jones Apr 05 '12 at 02:50

score 0 · Answer 2 · answered Apr 05 '12 at 02:41

0

you aren't mallocing space for all the structures themselves you have to add

for(int i = 0; i < 5; i++) {
    neato[i] = malloc(sizeof(testStruct));
}

After you malloc neato. Also you should check your return value from malloc for NULL to make sure malloc passed.

answered Apr 05 '12 at 02:41

twain249

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No need to cast the result of `malloc` in C. – Ed S. Apr 05 '12 at 02:42
@EdS. True but it doesn't hurt. – twain249 Apr 05 '12 at 02:44
It can actually. It's just totally unnecessary, serving only to add verbosity to your code, and it can actually be bad in a more meaningful way. See: http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – Ed S. Apr 05 '12 at 02:44

score 0 · Answer 3 · answered Apr 05 '12 at 02:42

0

You allocated array of pointers, but did not assign valid address to these pointers.

If you only want to create dynamic array, use just pointer to the struct:

testStruct *neato;
neato = malloc( sizeof(testStruct) * 5);
scanf("%d", &neato[0].test1);

answered Apr 05 '12 at 02:42

asaelr

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Pointer to a Pointer to a Structure and malloc memory for array

3 Answers3