保证复制省略的行为是否取决于用户定义的复制构造函数的存在? [英] Does the behavior of guaranteed copy elision depend on existence of user-defined copy constructor?

问题描述

在GCC 8.0.1下，无论有没有用户定义的复制构造函数，以下代码的行为都不同:

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);
}

使用任何一个注释的用户定义的副本构造函数(即使使用#2，也可以执行简单的浅表复制)，断言不会失败，这意味着

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.

但是，如果没有任何用户定义的副本构造函数，则断言将失败，这意味着main函数中的对象s不是默认构造的.为什么会这样?不能保证复制省略在这里执行吗?

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?

推荐答案

引用自C ++ 17工作草案§15.2临时对象第3段(

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

当将类型X的对象传递给函数或从函数中返回时，如果X的每个副本构造函数，移动构造函数和析构函数都是微不足道的或已删除，并且X至少具有一个未删除的副本或移动构造函数，则允许实现创建一个临时对象来保存功能参数或结果对象. ... [注意:已授予此纬度，以允许将类类型的对象传递到寄存器中的函数或从其中返回. — 尾注]

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]

在您的情况下，当我将复制和移动构造函数都设为默认设置时:

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

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

GCC和Clang的断言也失败了.请注意，隐式定义的构造函数很简单.

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

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