C array size given by variable

Question

I found some code today that confused me. It did something like this:

#include <stdio.h>

int main(int argc, char **argv) {
    int x = 5;
    int foo[x];

    foo[0] = 33;
    printf("%d\n", foo[0]);
    return 0;
}

My Question is why does this work?

The array foo is on the stack so how could it be expanded by x ?

I would have expected some thing like this:

#include <stdio.h>

int main(int argc, char **argv) {
    int x = 5;
    int foo[] = malloc(sizeof(int)*x);

    foo[0] = 33;
    printf("%d\n", foo[0]);
    free(foo);
    return 0;
}

Not that it is prettier or something but, I just wonder.

Answer 1

The snippet

int foo[x];

is talking advantage of something called VLA ( Variable length array ) feature. It was introduced in C99 standard, just to be made an optional feature in C11 .

This way, we can create an array data structure, whose length is given (supplied) at run-time.

Point to note, though created at runtime, gcc allocates the VLAs on stack memory (unlike the dynamic memory allocation from heap memory).

Answer 2

The array foo is on the stack so how could it be expanded by x?

gcc simply moves the stack pointer:

subq    %rax, %rsp

Link to full example with assembly output