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* [Bug c/114831] New: typeof doesn't evaluate expression when it has variably modified type in some cases
@ 2024-04-24 1:48 mforney at mforney dot org
0 siblings, 0 replies; only message in thread
From: mforney at mforney dot org @ 2024-04-24 1:48 UTC (permalink / raw)
To: gcc-bugs
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=114831
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 type in
some cases.
Consider the following program:
#include <stdio.h>
int n = 1;
int main(void) {
int a[n], (*p)[n] = 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
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2024-04-24 1:48 [Bug c/114831] New: typeof doesn't evaluate expression when it has variably modified type in some cases mforney at mforney dot org
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