为了演示移动语义,我使用int中的隐式构造函数编写了以下示例代码.
struct C {
int i_=0;
C() {}
C(int i) : i_( i ) {}
C( const C& other) :i_(other.i_) {
std::cout << "A copy construction was made." << i_<<std::endl;
}
C& operator=( const C& other) {
i_= other.i_ ;
std::cout << "A copy assign was made."<< i_<<std::endl;
return *this;
}
C( C&& other ) noexcept :i_( std::move(other.i_)) {
std::cout << "A move construction was made." << i_ << std::endl;
}
C& operator=( C&& other ) noexcept {
i_ = std::move(other.i_);
std::cout << "A move assign was made." << i_ << std::endl;
return *this;
}
};
和
auto vec2 = std::vector<C>{1,2,3,4,5};
cout << "reversing\n";
std::reverse(vec2.begin(),vec2.end());
随着输出
A copy construction was made.1 A copy construction was made.2 A copy construction was made.3 A copy construction was made.4 A copy construction was made.5 reversing A move construction was made.1 A move assign was made.5 A move assign was made.1 A move construction was made.2 A move assign was made.4 A move assign was made.2
现在,反过来显示了2个两个交换(每个交换使用一个移动分配和两个移动构造),但为什么从初始化列表创建的临时C对象无法移动？我以为我有一个整数的初始化列表,但我现在想知道我之间是否有一个Cs的初始化列表,它不能被移动(作为它的const).这是正确的解释吗？ – 这是怎么回事？
解决方法
I thought I had an initializer list of integers,but I’m now wondering if what I have in between is an initializer list of Cs,which can’t be moved from (as its const). Is this a correct interpretation?
这是对的.矢量< C>没有initializer_list< int>构造函数或甚至是initializer_list< T>一些模板参数T的构造函数.它具有的是initializer_list< C>构造函数 – 它是从你传入的所有int构建的.由于initializer_list的支持是一个const数组,你得到一堆副本而不是一堆移动.