How can I determine to which C standard is this code compliant?

Question

In a C lab this uber simple code appears:

#include <stdio.h>

int suma (int a, int b)
{
        return a+b;
}

int mult (int a, int b)
{
        return a*b;
}

int main(void)
{
        int a,b;
        printf ("Operando 1: ");
        scanf("%d",&a);
        printf("Operando 2: ");
        scanf("%d",&b);
        printf("%d+%d=%d\n",a,b,suma(a,b));
        printf("%d*%d=%d\n",a,b,mult(a,b));
        return 0; 
}

by looking at the code I'm supposed to determine to which C standard it is compliant (ANSI, ISO or de facto K&R). After reading this and this I'm inclined to say that it's compliant to the three standards. Would that be correct?

Answer 1

If you happen to use gcc, with the -std=..., -Wall and -pedantic options you may change the standard that it expects and get warning / errors if the code doesn't comply.

Answer 2

K&R C does not include arguments in function declarations.

See, for instance, the Wikipedia text on C's history.

Answer 3

It's not K&R. K&R function declarations define arguments outside the parentheses like this:

int mult (a, b)
    int a;
    int b;

and void was ntroduced with the first ANSI standard.

To me it looks likeit is compliant with both ANSI C89 and C99.

Answer 4

I say C89 (ANSI C89 or ISO C90).

It isn't pre-standard (K&R) C: function definition has type information for parameters inside the declaration.

It follows all rules of the C89 standard and all rules of the C99 standard (*). However, the C99 standard says that main returns 0 if it reaches the closing } without a return statement. So, the guy who wrote this had to add the return 0; statement ... or the question doesn't make much sense.

(*) the first 2 printf shoud really have a '\n' (or call fflush(stdout); afterwards) to flush the output

Answer 5

#if __STDC_VERSION__ >= 199901L
puts("C99");
#else
puts("C89 or pre C89");
#endif

will work for ANSI C compilers.