Variable-length arrays in C89?

Question

I've read that C89 does not support variable-length arrays, but the following experiment seems to disprove that:

#include <stdio.h>

int main()
{
   int x;
   printf("Enter a number: ");
   scanf("%d", &x);
   int a[x];
   a[0] = 1;
   // ...
   return 0;
}

When I compile as such (assuming filename is va_test.c ):

gcc va_test.c -std=c89 -o va_test

It works...

What am I missing? :-)

Answer 1

GCC always supported variable length arrays AFAIK. Setting -std to C89 doesn't turn off GCC extensions ...

Edit: In fact if you check here:

http://gcc.gnu.org/onlinedocs/gcc/C-Dialect-Options.html#C-Dialect-Options

Under -std= you will find the following:

ISO C90 programs ( certain GNU extensions that conflict with ISO C90 are disabled). Same as -ansi for C code.

Pay close attention to the word "certain".

Answer 2

C89 does not recognize // comments.

C89 does not allow definitions intermixed with code.

You need to fflush(stdout) after the printf to be sure of seing the prompt before the scanf .

main "looks better" as int main(void)

Try gcc -std=c89 -pedantic ... instead

Answer 3

You're missing that without -pedantic , gcc isn't (and doesn't claim to be) a standard-conforming C compiler. Instead, it compiles a GNU dialect of C, which includes various extensions.