上下文:我正在进行实验,以了解gcc何时执行RVO,如果不执行,何时使用移动语义。我的gcc版本是g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-4)

问题:我有一个按值返回Foo的函数。编译器无法执行RVO,因为有两个可能的命名返回值。当我使用三元运算符选择要返回的Foo时,则需要显式调用std::move以避免复制。使用if语句时,不需要std::move。为什么会有差异?

码:

#include <iostream>

using namespace std;

struct Foo {
    std::string s;
    Foo()                                        { cout << "Foo()\n"; }
    ~Foo()                                       { cout << "~Foo()\n"; }
    Foo(const Foo& other)     : s(other.s)       { cout << "Foo(const Foo&)\n"; }
    Foo(Foo&& other) noexcept : s(move(other.s)) { cout << "Foo(Foo&&)\n"; }
};

Foo makeFooIf(bool which) {
    Foo foo1; foo1.s = "Hello, World1!";
    Foo foo2; foo2.s = "Hello, World2!";
    if (which) return foo1;
    else       return foo2;
}

Foo makeFooTernary(bool which) {
    Foo foo1; foo1.s = "Hello, World1!";
    Foo foo2; foo2.s = "Hello, World2!";
    return which ? foo1 : foo2;
}

Foo makeFooTernaryMove(bool which) {
    Foo foo1; foo1.s = "Hello, World1!";
    Foo foo2; foo2.s = "Hello, World2!";
    return which ? move(foo1) : move(foo2);
}

int main()
{
    cout << "----- makeFooIf -----\n";
    Foo fooIf = makeFooIf(true);
    cout << fooIf.s << endl;

    cout << "\n----- makeFooTernary -----\n";
    Foo fooTernary = makeFooTernary(true);
    cout << fooTernary.s << endl;

    cout << "\n----- makeFooTernaryMove -----\n";
    Foo fooTernaryMove = makeFooTernaryMove(true);
    cout << fooTernaryMove.s << endl;

    cout << "\n----- Cleanup -----\n";
    return 0;
}

输出:
----- makeFooIf -----
Foo()
Foo()
Foo(Foo&&)
~Foo()
~Foo()
Hello, World1!

----- makeFooTernary -----
Foo()
Foo()
Foo(const Foo&)
~Foo()
~Foo()
Hello, World1!

----- makeFooTernaryMove -----
Foo()
Foo()
Foo(Foo&&)
~Foo()
~Foo()
Hello, World1!

----- Cleanup -----
~Foo()
~Foo()
~Foo()

最佳答案

在某些情况下会有一个隐含的举动:

§12.8.32



我的大胆

关于c++ - 使用if语句 move 构造函数,但使用三元运算符复制构造函数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46107480/

10-11 22:51
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