CODEWITHSUNDEEP
carrays
maxbit89
maxbit89

Reputation: 770

C array size given by variable

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.

Upvotes: 4

Views: 199

Answers (2)

M.M
M.M

Reputation: 141648

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

Upvotes: 2

Sourav Ghosh
Sourav Ghosh

Reputation: 134386

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).

Upvotes: 9

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