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? :-)
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".
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
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.
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