Why is variable-length array working in this C code?

Question

#include "stdio.h"

int main()
    {
        int n;
        printf("Enter n:\n");
        scanf("%d",&n);

        int arr[n][n];
        arr[3][3] = 4;
        printf("%d",arr[3][3]);
        getchar();
        return 0;
    }

Wasn't using int arr[n], where n is a variable, illegal in C? I am trying to understand what's happening here. Apparently the code works on my clang LLVM compiler and on IDEOne and on Codeblocks. I guess the compiler is just making things easy for me, by doing automatic memory allocation. But another astounding fact is that when I try to set n to 1, 2 or 3 it still works.

They're illegal in ANSI C. They were added to the standard in C99. — Chris Hayes, Dec 28 '13 at 08:19

score 6 · Answer 1 · answered Dec 28 '13 at 08:20

6

Variable length arrays are allowed by C standard since C99. Note that they are still not allowed by C++
standard.
Also important point to note is for versions before c99 and for most C++ compilers variable length arrays are supported by implementations in the form of extensions.

answered Dec 28 '13 at 08:20

Alok Save

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@AneeshDogra Because you're invoking [undefined behavior](http://stackoverflow.com/questions/15646973/how-dangerous-is-it-to-access-an-array-out-of-bounds) and got lucky (or unlucky depending on how you look at it). – Chris Hayes Dec 28 '13 at 08:22
In fact, 3 is the same case. Your example is legal with `n >= 4`. – el.pescado - нет войне Dec 28 '13 at 08:48

Why is variable-length array working in this C code?

1 Answers1