移动构造函数不被调用 [英] move Constructor is not called

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本文介绍了移动构造函数不被调用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在实现一个IntArray类来学习C ++.我必须承认,我还没有完全理解r和lvalues和移动构造函数.我想尝试一下,看看我的代码是否正常工作,但是我不知道为什么 {IntArray array = IntArray(5);}不会调用我实现的move构造函数.我以为这是一个例子.

I am implementing a IntArray Class for learning C++. I must admit I haven't fully understood r and lvalues and move constructors, yet. I wanted to try it out to see if my code is working, but I do not know why {IntArray array = IntArray(5);} doesn't call my implemented move constructor. I thought this would be a case for it.

#include "IntArray.h"
IntArray::IntArray()
    :data(nullptr), count(0), capacity(0) {std::cout << "Default Constructor called" << std::endl;}

IntArray::IntArray(int size)
    :data(new int[size]), count(size), capacity(size)  {std::cout << "Constructor called with size: " << size << std::endl;}

IntArray::~IntArray() {
    std::cout << "Destructor called" << std::endl;
    delete[] data; //rest is stack allocated and gets freed with end of scope
}

//COPY CONSTRUCT & ASSIGN OP
IntArray::IntArray(const IntArray& rhs) 
    :data(new int[rhs.count]), count(rhs.count), capacity(rhs.count) //warum nicht mit capacity? wir wollen doch eine exakte kopie?
{
    std::cout << "Copy Constructor called" << std::endl;
    std::copy(rhs.data, rhs.data + rhs.count, data); //pointer arithmetik?
}
IntArray& IntArray::operator=(const IntArray& rhs) {
    if (&rhs == this) //check for selfassign
        return *this;

    //if capacity of lhs is NOT big enough, reallocate new
    if (capacity < rhs.capacity) { 
        delete[] data;
        data = new int[rhs.count];
        capacity = rhs.count;
    }   
    count = rhs.count;
    std::copy(rhs.data, rhs.data + rhs.count, data);
    return *this;
}

//MOVE CONSTRUCT & ASSIGN OP
IntArray::IntArray(IntArray&& rhs)
    :data(rhs.data), count(rhs.count), capacity(rhs.capacity)
{
    std::cout << "Move Constructor called" << std::endl;

    rhs.data = nullptr;
    rhs.count = 0;
    rhs.capacity = 0;
}
IntArray& IntArray::operator=(IntArray&& rhs) {
    if (&rhs == this) //self assignment?
        return *this;
    std::cout << "Move assignment operator called" << std::endl;

    //steal
    delete[] data;
    data = rhs.data;
    count = rhs.count;
    capacity = rhs.capacity;

    //Reset old obj to prevent double freeing
    rhs.data = nullptr;
    rhs.count = 0;
    rhs.capacity = 0;

    return *this;
}

推荐答案

您看不到移动构造或移动分配仅仅是优化的结果!请查看复制省略" https://en.cppreference.com/w /cpp/language/copy_elision

That you don't see move construction or move assignment is simply a result of optimization! Please take a look for "copy elision" https://en.cppreference.com/w/cpp/language/copy_elision

如果您使用的是gcc,则可以通过以下方式告诉编译器不要对其进行优化:

If you are using gcc you can tell the compiler to don't optimize it by:

g++ -O0 main.cpp -fno-elide-constructors

现在的结果是:

Constructor called with size: 5
Move Constructor called
Destructor called
Destructor called

这篇关于移动构造函数不被调用的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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