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
– Chris Hayes, Commented Dec 28, 2013 at 8:19

Alok Save · Accepted Answer · 2013-12-28 08:20:30Z

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, 2013 at 8:20

Alok Save

207k56 gold badges448 silver badges539 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Chris Hayes Over a year ago

@AneeshDogra Because you're invoking undefined behavior and got lucky (or unlucky depending on how you look at it).

el.pescado - нет войне Over a year ago

In fact, 3 is the same case. Your example is legal with n >= 4.

Collectives™ on Stack Overflow

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

1 Answer 1

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related