Re: case label does not reduce to an integer constant


"Rex Mottram" <rexm@xxxxxxxx> wrote in message
news:5JmdnaLkgpQwPJ3VnZ2dnUVZ_uKpnZ2d@xxxxxxxxxxxxxx
Can anyone explain this?

% cat t.c
#include <stdio.h>

#define FIRST "first"
#define SECOND "second"

int
main(int argc, char *argv[])
{
switch (argv[1][0]) {
case FIRST[0]:
case 'f':
printf("%s == FIRST\n", argv[1]);
break;
case SECOND[0]:
case 's':
printf("%s == SECOND\n", argv[1]);
break;
}
return 0;
}

% gcc -o t -W -Wall -g t.c
t.c: In function `main':
t.c:10: error: case label does not reduce to an integer constant
t.c:14: error: case label does not reduce to an integer constant

It certainly looks constant to me, and it should be promotable to an
integer exactly as the neighboring case label is. This is gcc 3.4.5 FWIW.

Possibly not allowed for the same reason that:

int x[FIRST[0]];

is not allowed, when declared outside a function (inside a function, it
might work as a VLA, but it is still not considered a constant).

I suppose there is a limit to what a compiler can do to determine an
expression is constant, or maybe it considers that the string "first" could
change (I managed to do this on lccwin, but gcc generated a crash when
attempting to change the constant string).

--
Bart


.

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