Why can std::function be constructed with a lambda

问题:

The following compiles fine:

#include <functional>

int main()
{
    std::function<const int&()> f = []() -> int {return 1;};
    const int& r = f(); // r is a dangling reference
    return 0;
}

How come it's possible to set an std::function with a const int& return type to a lambda with an int return type? Allowing this sort of cast to happen implicitly and with no warning is a gotcha IMHO.

回答1:

You can construct a std::function with any object which is callable with the relevant arguments and whose return value is implicitly convertible to the std::function return. int is implicitly convertible to const int&, so the rules are met.

A compiler could feel free to warn about this, but it seems like a lot of work for a particularly corner-y corner case.