C++ 11移动语义

Question

我试图了解C++ 11移动语义是如何工作的。我已经实现了一个类，它包装了指向字符串对象的指针，但既没有按预期调用移动构造函数，也没有调用移动赋值操作符。

我通过Eclipse使用GCC 4.7.2：

你能帮我理解原因吗？

#include <iostream>
#include <string>
#include <utility>

using namespace std;

class StringPointerWrapper {
public:

    // Default constructor with default value
    StringPointerWrapper(const std::string& s = "Empty"): ps(new std::string(s)) {
        std::cout << "Default constructor: " << *ps << std::endl;
    }

    //Copy constructor
    StringPointerWrapper(const StringPointerWrapper& other): ps(new std::string(*other.ps)) {
        std::cout << "Copy constructor: " << *other.ps << std::endl;
    }

    //Copy assignment operator
    StringPointerWrapper& operator=(StringPointerWrapper other) {
        std::cout << "Assignment operator (ref): " << *other.ps << std::endl;
        swap(ps, other.ps);
        return *this;
    }


    //Alternate copy assignment operator
    /*StringPointerWrapper& operator=(StringPointerWrapper& other) {
        std::cout << "Assignment operator (val)" << std::endl;
        //We need to do the copy by ourself
        StringPointerWrapper temp(other);
        swap(ps, temp.ps);
        return *this;
    }*/

    //Move constructor
    StringPointerWrapper(StringPointerWrapper&& other) noexcept : ps(nullptr) {
        std::cout << "Move constructor: " << *other.ps << std::endl;
        ps = other.ps;
        other.ps = nullptr;
    }

    //Move assignment operator
    StringPointerWrapper& operator= (StringPointerWrapper&& other) noexcept {
        std::cout << "Move assignment operator: " << *other.ps << std::endl;
        if(this != &other) {
            delete ps;
            ps = other.ps;
            other.ps = nullptr;
        }
        return *this;
    }


    //Destructor
    ~StringPointerWrapper() {
        std::cout << "Destroying: " << *this << std::endl;
        delete ps;
    }
private:
        friend std::ostream& operator<<(std::ostream& os, StringPointerWrapper& spw) {
            os << *spw.ps;
            return os;
        }
        std::string *ps;
};


int main(int argc, char *argv[]) {

    StringPointerWrapper spw1("This is a string");
    StringPointerWrapper spw2;
    StringPointerWrapper spw3("This is another string");
    StringPointerWrapper spw4 = {"This is a const string"};
    StringPointerWrapper spw5(StringPointerWrapper("String for move constructor"));
    std::cout << "spw2 before: " << spw2 << std::endl;
    spw2 = spw3;
    std::cout << "spw2 after: " << spw2 << std::endl;
    StringPointerWrapper spw6 = StringPointerWrapper("String for move assignment");
    std::cout << spw1 << std::endl;
    std::cout << spw2 << std::endl;
    std::cout << spw3 << std::endl;
    std::cout << spw4 << std::endl;
    std::cout << spw5 << std::endl;
    std::cout << spw6 << std::endl;
}

Answer 1

未调用move构造函数，因为编译器正在优化构造函数。如果在编译器调用中传递-fno-elide-constructors，则可以禁用它。

但是，您有问题，因为您的move构造函数只是使用来自other的指针，该指针很快就会被删除。将std::string作为指针保存是没有意义的，您应该直接持有它，并在移动赋值操作符和移动构造函数中调用std::move。

Answer 2

StringPointerWrapper spw5(StringPointerWrapper(用于移动构造函数的字符串))；

这会调用默认构造函数，因为编译器已决定通过不创建临时构造函数(即，它已决定执行StringPointerWrapper spw5("String for move constructor"))来优化该构造函数。相反，通过执行StringPointerWrapper spw5(std::move(StringPointerWrapper("String for move constructor")));强制移动。

StringPointerWrapper spw6 =StringPointerWrapper(“用于移动分配的字符串”)；

同样，编译器通过优化临时创建来调用默认构造函数。

注意：您的operator<<需要防范空指针。例如,

friend std::ostream& operator<<(std::ostream& os, StringPointerWrapper& spw) {
    if (spw.ps)
         os << (spw.ps);
        else
            os << "null";
    return os;
 }

Answer 3

您需要自己告诉编译器用std::move移动东西

尝试：

 StringPointerWrapper spw5(std::move(StringPointerWrapper("String for move constructor")));