https://gcc.gnu.org/bugzilla/show_bug.cgi?id=104803
Bug ID: 104803
Summary: if consteval error from branch that isn't evaluated
anyway
Product: gcc
Version: 12.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: barry.revzin at gmail dot com
Target Milestone: ---
Consider this example. Yes, the body of none_of here looks... extremely
bizarre. Just bear with me.
template <typename R, typename P>
constexpr auto none_of(R&& r, P p) -> bool {
for (int i : r) {
if consteval {
if (p(i)) { // <== line 5
return false;
}
} else {
if (p(i)) { // <== line 9
return false;
}
}
}
return true;
}
constexpr int vals[] = {1, 0, -1};
static_assert(none_of(vals,
[](int i) consteval {
return i > 2;
}));
This does not compile, with the error (https://godbolt.org/z/dEq3381ac):
<source>: In instantiation of 'constexpr bool none_of(R&&, P) [with R = const
int (&)[3]; P = <lambda(int)>]':
<source>:18:22: required from here
<source>:9:19: error: the value of 'i' is not usable in a constant expression
9 | if (p(i)) {
| ^
<source>:3:14: note: 'int i' is not const
3 | for (int i : r) {
| ^
Compiler returned: 1
The error is on line 9. That's the second part of the if consteval, the "else"
part. That's not even supposed to be evaluated here, since we're in a
static_assert. If I replace it with 'return false':
template <typename R, typename P>
constexpr auto none_of(R&& r, P p) -> bool {
for (int i : r) {
if consteval {
if (p(i)) {
return false;
}
} else {
return false;
}
}
return true;
}
constexpr int vals[] = {1, 0, -1};
static_assert(none_of(vals,
[](int i) consteval {
return i > 2;
}));
This compiles (https://godbolt.org/z/od93dEMf8). If 'return false;' in the else
branch were executed, this would've failed. But it's (correctly) not.
Something about the if consteval machinery is incorrectly evaluating the wrong
branch and failing as a result.
