clang - shared_ptr fails to run its deleter - c++

Question

This code prints 0 (with no optimization) or 666 (with optimizations turned on) when built with clang++ -std=c++11 (-O3 yields 666, which is what I would expect). When lambda is passed by universal reference the problem disappears.
FYI, GCC prints 666 on all the versions I've tested.
Is it a compiler bug or the code is incorrect?
#include <memory>
#include <iostream>
template <typename T>
std::shared_ptr<void> onScopeExit(T f)
{
return std::shared_ptr<void>((void*)1, [&](void *) {
f();
});
}
struct A {
void f() {
auto scopeGuard = onScopeExit([&]() { i = 666; }); // [1]
// ... (some work)
} // (lambda [1] being ? called on scope exit)
int i = 0;
};
A a;
int main() {
a.f();
std::cout << a.i << std::endl;
}
The compiler in question is:
Apple LLVM version 9.1.0 (clang-902.0.39.2)
Target: x86_64-apple-darwin17.7.0

Answer 1

Your code has undefined behavior. You capture f by reference in onScopeExit but once you return the shared_ptr from the function the deleter is now holding a dangling reference to f, since f went out of scope. What you need to do is capture f by value, and then you won't have a dangling reference
template <typename T>
std::shared_ptr<void> onScopeExit(T f)
{
return std::shared_ptr<void>((void*)1, [=](void *) {
f();
});
}
struct A {
void f() {
auto scopeGuard = onScopeExit([&]() { i = 666; }); // [1]
// ... (some work)
} // (lambda [1] being ? called on scope exit)
int i = 0;
};
A a;
int main() {
a.f();
std::cout << a.i << std::endl;
}