Tyker added a comment.
The now that constexpr destructors are legal. The code in this patch need to be
adapted, I have question about the following code.
struct A {
constexpr ~A() {}
};
consteval A f() {
return A{};
}
void test() {
A a;
a = f(); // <-- here
}
At the point i marked.
The invocation of f causes an immediate invocation
(http://eel.is/c++draft/expr.const#12).
Immediate invocation are full expression
(http://eel.is/c++draft/intro.execution#5).
Full expression resolve all there side-effects before evaluating the next full
expression (http://eel.is/c++draft/intro.execution#9).
The return value of f() is created inside the immediate invocation.
So the destructor of the value returned by f() should be destroyed within the
immediate evaluation.
But that value is needed for the assignment operator.
This seem contradictory. What have i misunderstood ? What should happen here ?
CHANGES SINCE LAST ACTION
https://reviews.llvm.org/D63960/new/
https://reviews.llvm.org/D63960
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