按值传递的意义是什么？
当一个函数的参数按值传递时，这就会进行拷贝。当然，编译器懂得如何去拷贝。
而对于我们自定义的类型，我们也许需要提供拷贝构造函数。
但是不得不说，拷贝的代价是昂贵的。
所以我们需要寻找一个避免不必要拷贝的方法，即C++11提供的移动语义。
上一篇博客中有一个句话用到了：
#include <iostream>
void f(int& i) { std::cout << "lvalue ref: " << i << "\n"; }
void f(int&& i) { std::cout << "rvalue ref: " << i << "\n"; }
int main()
{
int i = 77;
f(i); // lvalue ref called
f(99); // rvalue ref called
f(std::move(i)); // 稍后介绍
return 0;
}实际上，右值引用注意用于创建移动构造函数和移动赋值运算。
移动构造函数类似于拷贝构造函数，把类的实例对象作为参数，并创建一个新的实例对象。
但是 移动构造函数可以避免内存的重新分配，因为我们知道右值引用提供了一个暂时的对象，而不是进行copy，所以我们可以进行移动。
换言之，在设计到关于临时对象时，右值引用和移动语义允许我们避免不必要的拷贝。我们不想拷贝将要消失的临时对象，所以这个临时对象的资源可以被我们用作于其他的对象。
右值就是典型的临时变量，并且他们可以被修改。如果我们知道一个函数的参数是一个右值，我们可以把它当做一个临时存储。这就意味着我们要移动而不是拷贝右值参数的内容。这就会节省很多的空间。
说多无语，看代码：
#include <iostream>
#include <algorithm>
class A
{
public:
// Simple constructor that initializes the resource.
explicit A(size_t length)
: mLength(length), mData(new int[length])
{
std::cout << "A(size_t). length = "
<< mLength << "." << std::endl;
}
// Destructor.
~A()
{
std::cout << "~A(). length = " << mLength << ".";
if (mData != NULL) {
std::cout << " Deleting resource.";
delete[] mData; // Delete the resource.
}
std::cout << std::endl;
}
// Copy constructor.
A(const A& other)
: mLength(other.mLength), mData(new int[other.mLength])
{
std::cout << "A(const A&). length = "
<< other.mLength << ". Copying resource." << std::endl;
std::copy(other.mData, other.mData + mLength, mData);
}
// Copy assignment operator.
A& operator=(const A& other)
{
std::cout << "operator=(const A&). length = "
<< other.mLength << ". Copying resource." << std::endl;
if (this != &other) {
delete[] mData; // Free the existing resource.
mLength = other.mLength;
mData = new int[mLength];
std::copy(other.mData, other.mData + mLength, mData);
}
return *this;
}
// Move constructor.
A(A&& other) : mData(NULL), mLength(0)
{
std::cout << "A(A&&). length = "
<< other.mLength << ". Moving resource.\n";
// Copy the data pointer and its length from the
// source object.
mData = other.mData;
mLength = other.mLength;
// Release the data pointer from the source object so that
// the destructor does not free the memory multiple times.
other.mData = NULL;
other.mLength = 0;
}
// Move assignment operator.
A& operator=(A&& other)
{
std::cout << "operator=(A&&). length = "
<< other.mLength << "." << std::endl;
if (this != &other) {
// Free the existing resource.
delete[] mData;
// Copy the data pointer and its length from the
// source object.
mData = other.mData;
mLength = other.mLength;
// Release the data pointer from the source object so that
// the destructor does not free the memory multiple times.
other.mData = NULL;
other.mLength = 0;
}
return *this;
}
// Retrieves the length of the data resource.
size_t Length() const
{
return mLength;
}
private:
size_t mLength; // The length of the resource.
int* mData; // The resource.
};移动构造函数
语法：
A(A&& other) noexcept // C++11 - specifying non-exception throwing functions
{
mData = other.mData; // shallow copy or referential copy
other.mData = nullptr;
}最主要的是没有用到新的资源，是移动而不是拷贝。
假设一个地址指向了一个有一百万个int元素的数组，使用move构造函数，我们没有创造什么，所以代价很低。
// Move constructor.
A(A&& other) : mData(NULL), mLength(0)
{
// Copy the data pointer and its length from the
// source object.
mData = other.mData;
mLength = other.mLength;
// Release the data pointer from the source object so that
// the destructor does not free the memory multiple times.
other.mData = NULL;
other.mLength = 0;
}移动比拷贝更快！！！
移动赋值运算符
语法：
A& operator=(A&& other) noexcept
{
mData = other.mData;
other.mData = nullptr;
return *this;
}工作流程这样的：Google上这么说的：
Release any resources that *this currently owns.
Pilfer other’s resource.
Set other to a default state.
Return *this.
// Move assignment operator.
A& operator=(A&& other)
{
std::cout << "operator=(A&&). length = "
<< other.mLength << "." << std::endl;
if (this != &other) {
// Free the existing resource.
delete[] mData;
// Copy the data pointer and its length from the
// source object.
mData = other.mData;
mLength = other.mLength;
// Release the data pointer from the source object so that
// the destructor does not free the memory multiple times.
other.mData = NULL;
other.mLength = 0;
}
return *this;
}让我们看几个move带来的好处吧！
vector众所周知，C++11后对vector也进行了一些优化。例如vector::push_back()被定义为了两种版本的重载，一个是cosnt T&左值作为参数，一个是T&&右值作为参数。例如下面的代码：
std::vector<A> v;
v.push_back(A(25));
v.push_back(A(75));上面两个push_back()都会调用push_back(T&&)版本，因为他们的参数为右值。这样提高了效率。
而 当参数为左值的时候，会调用push_back(const T&) 。
#include <vector>
int main()
{
std::vector<A> v;
A aObj(25); // lvalue
v.push_back(aObj); // push_back(const T&)
}但事实我们可以使用 static_cast进行强制：
// calls push_back(T&&)
v.push_back(static_cast<A&&>(aObj));我们可以使用std::move完成上面的任务：
v.push_back(std::move(aObj)); //calls push_back(T&&)似乎push_back(T&&)永远是最佳选择，但是一定要记住：
push_back(T&&) 使得参数为空。如果我们想要保留参数的值，我们这个时候需要使用拷贝，而不是移动。
最后写一个例子，看看如何使用move来交换两个对象：
#include <iostream>
using namespace std;
class A
{
public:
// constructor
explicit A(size_t length)
: mLength(length), mData(new int[length]) {}
// move constructor
A(A&& other)
{
mData = other.mData;
mLength = other.mLength;
other.mData = nullptr;
other.mLength = 0;
}
// move assignment
A& operator=(A&& other) noexcept
{
mData = other.mData;
mLength = other.mLength;
other.mData = nullptr;
other.mLength = 0;
return *this;
}
size_t getLength() { return mLength; }
void swap(A& other)
{
A temp = move(other);
other = move(*this);
*this = move(temp);
}
int* get_mData() { return mData; }
private:
int *mData;
size_t mLength;
};
int main()
{
A a(11), b(22);
cout << a.getLength() << ‘ ‘ << b.getLength() << endl;
cout << a.get_mData() << ‘ ‘ << b.get_mData() << endl;
swap(a,b);
cout << a.getLength() << ‘ ‘ << b.getLength() << endl;
cout << a.get_mData() << ‘ ‘ << b.get_mData() << endl;
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
时间： 2024-10-06 02:08:47