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From: "mforney at mforney dot org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug c/114831] New: typeof doesn't evaluate expression when it has
 variably modified type in some cases
Date: Wed, 24 Apr 2024 01:48:58 +0000
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X-Bugzilla-Product: gcc
X-Bugzilla-Component: c
X-Bugzilla-Version: 14.0
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D114831

            Bug ID: 114831
           Summary: typeof doesn't evaluate expression when it has
                    variably modified type in some cases
           Product: gcc
           Version: 14.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c
          Assignee: unassigned at gcc dot gnu.org
          Reporter: mforney at mforney dot org
  Target Milestone: ---

C23 introduced typeof, and 6.7.2.5p4 says

> If the type of the operand is a variably modified type, the operand
> is evaluated; otherwise, the operand is not evaluated.

However, gcc doesn't evaluate the operand when it has variably modified typ=
e in
some cases.

Consider the following program:

#include <stdio.h>
int n =3D 1;
int main(void) {
        int a[n], (*p)[n] =3D 0;
        typeof(puts("&a is variably-modified"), &a) p1;
        typeof(puts("p is variably-modified"), p) p2;
        typeof(puts("p1 is variably-modified"), p1) p3;
        typeof(puts("p2 is variably-modified"), p2) p4;
}

All four of the typeof operands have the same type, int (*)[n]. But when I =
run
this program it prints just:

p is variably-modified
p2 is variably-modified=