Does the behavior of guaranteed copy elision depend on existence of user-defined copy constructor?

半世苍凉 提交于 2019-12-03 08:54:04

问题


The following code behaves differently with or without user-defined copy constructor under GCC 8.0.1:

#include <cassert>

struct S {
    int i;
    int *p;
    S() : i(0), p(&i) {}
    // S(const S &s) : i(s.i), p(&i) {}  // #1
    // S(const S &s) : i(s.i), p(s.p) {} // #2
    // S(const S &s) = delete;           // #3
};

S make_S() {return S{};}

int main()
{
    S s = make_S();
    assert(s.p == &s.i);
}

With either of the commented user-defined copy constructors (even with #2, the one performing a simple shallow copy), the assertion will not fail, which means guaranteed copy elision works as expected.

However, without any user-defined copy constructor, the assertion fails, which means the object s in main function is not default-constructed. Why does this happen? Doesn't guaranteed copy elision perform here?


回答1:


Quoting from C++17 Working Draft §15.2 Temporary Objects Paragraph 3 (https://timsong-cpp.github.io/cppwp/class.temporary#3):

When an object of class type X is passed to or returned from a function, if each copy constructor, move constructor, and destructor of X is either trivial or deleted, and X has at least one non-deleted copy or move constructor, implementations are permitted to create a temporary object to hold the function parameter or result object. ... [ Note: This latitude is granted to allow objects of class type to be passed to or returned from functions in registers. — end note]

In your case, when I made both copy and move constructors defaulted:

S(const S &) = default;
S(S &&) = default;

assertion failed as well with GCC and Clang. Note that implicitly-defined constructors are trivial.



来源:https://stackoverflow.com/questions/48879226/does-the-behavior-of-guaranteed-copy-elision-depend-on-existence-of-user-defined

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