c++ - 复制省略的条件？

Question

我想验证以下优化是否按预期工作:

• RVO
• 命名 RVO
• 按值传递参数时复制省略

所以我写了这个小程序:

#include <algorithm>
#include <cstddef>
#include <iostream>
#include <vector>

struct Foo {
    Foo(std::size_t length, char value) : data(length, value) { }

    Foo(const Foo & rhs) : data(rhs.data) { std::cout << "*** COPY ***" << std::endl; }

    Foo & operator= (Foo rhs) {
        std::cout << "*** ASSIGNMENT ***" << std::endl;
        std::swap(data, rhs.data); // probably expensive, ignore this please
        return *this;
    }

    ~Foo() { }

    std::vector<char> data;
};

Foo TestRVO() { return Foo(512, 'r'); }

Foo TestNamedRVO() { Foo result(512, 'n'); return result; }

void PassByValue(Foo inFoo) {}

int main()
{
    std::cout << "\nTest RVO: " << std::endl;
    Foo rvo = TestRVO();

    std::cout << "\nTest named RVO: " << std::endl;
    Foo named_rvo = TestNamedRVO();

    std::cout << "\nTest PassByValue: " << std::endl;
    Foo foo(512, 'a');
    PassByValue(foo);

    std::cout << "\nTest assignment: " << std::endl;
    Foo f(512, 'f');
    Foo g(512, 'g');
    f = g;
}

我在启用优化的情况下编译了它:

$ g++ -o test -O3 main.cpp ; ./test

这是输出:

Test RVO: 

Test named RVO: 

Test PassByValue: 
*** COPY ***

Test assignment: 
*** COPY ***
*** ASSIGNMENT ***

根据输出 RVO 和命名 RVO 按预期工作。但是，赋值运算符和调用 PassByValue 时不会执行复制省略。

用户定义的复制构造函数是否不允许复制省略？ (我知道标准明确允许 RVO，但我不知道按值传递时的复制省略。)有没有一种方法可以在不定义复制构造函数的情况下验证复制省略？

Answer 1

标准说(在第 12.8.15 段中):

This elision of copy operations is permitted in the following circumstances (which may be combined to eliminate multiple copies):

• in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object with the same cv-unqualified type as the function return type, the copy operation can be omitted by constructing the automatic object directly into the function’s return value

• when a temporary class object that has not been bound to a reference (12.2) would be copied to a class object with the same cv-unqualified type, the copy operation can be omitted by constructing the tempo- rary object directly into the target of the omitted copy

这两种情况都不适用于此处，因此不允许省略。第一个很明显(不返回)。第二种是不允许的，因为你传入的对象不是临时的。

请注意，您的代码仍然没有问题，因为无论如何您都必须创建拷贝。要删除该拷贝，您将不得不使用 C++0x 的移动语义。