Why am I getting segmentation fault for malloc() while using pointer to pointer?

Question

I don't understand why this works:

void main() { 
    int *  b;
    b = (int *)malloc(sizeof(int));
    *b = 1;
    printf("*b = %d\n", *b);
}

while this does not (gets segmentation fault for the malloc()):

void main() {

    int ** a;
    int    i;

    for (i = 0; i<= 3; i++) {
        a[i] = (int*)malloc(sizeof(int));
        *(a[i]) = i;
        printf("*a[%d] = %d\n", i, *(a[i]));
    }
}

since I find a[i] is just like b in the first example.

BTW, a[i] is equal to *(a+i), right?

Use `int main(void)` henceforth.`void main()` is not portable. — Rüppell's Vulture, May 15 '13 at 07:25
Also [never cast the result of `malloc`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc/605858#605858). — Paul R, May 15 '13 at 07:27

score 3 · Accepted Answer · answered May 15 '13 at 07:27

3

You need to allocate memory for a first, so that you can access its members as a[i].

So if you want to allocate for 4 int * do

a = malloc(sizeof(int *) * 4);

for (i = 0; i<= 3; i++) {
  ...
}

or define it as array of integer pointers as

int *a[4];

answered May 15 '13 at 07:27

Rohan

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@Rüppell'sVulture, not sure what are you asking. – Rohan May 15 '13 at 08:00
Nothing,there was a misunderstanding about the `.....` part.I thought you had supplanted the `malloc()` segment to outside the loop.But it works fine when we use `int **a = malloc(sizeof(int *) * 4);` and leave the rest of the program intact.Maybe that's what you meant and I misread it. – Rüppell's Vulture May 15 '13 at 08:02

score 1 · Answer 2 · answered May 15 '13 at 07:34

a is a 2 dimensional pointer, you have to allocate both dimension. b is a 1 dimensional pointer, you have to allocate only one dimension and that's what you're doing with

b = (int *)malloc(sizeof(int));

So in order the second example to work you have to allocate the space for the pointer of pointer

void main() {

int ** a;
int    i;
a = (int**)malloc(4*sizeof(int*));

for (i = 0; i<= 3; i++) {
    a[i] = (int*)malloc(sizeof(int));
    *(a[i]) = i;
    printf("*a[%d] = %d\n", i, *(a[i]));
}

score 0 · Answer 3 · edited May 23 '17 at 11:55

The allocated pointer is written to uninitialized memory (you never set a to anything), causing undefined behavior.

So no, it's not at all equivalent to the code in the first example.

You would need something like:

int **a;

a = malloc(3 * sizeof *a);

first, to make sure a holds something valid, then you can use indexing and assign to a[0].

Further, this:

a[i] = (int*)malloc(sizeof(int));

doesn't make any sense. It's assigning to a[i], an object of type int *, but allocating space for sizeof (int).

Finally, don't cast the return value of malloc() in C.

score 0 · Answer 4 · answered May 15 '13 at 07:33

actually malloc it's not that trivial if you really want safe and portable, on linux for example malloc could return a positive response for a given request even if the actual memory it's not even really reserved for your program or the memory it's not writable.

For what I know both of your examples can potentially return a seg-fault or simply crash.

@ruppells-vulture I would argue that malloc is really portable and "safe" for this reasons.

Why am I getting segmentation fault for malloc() while using pointer to pointer?

4 Answers4