[英]C++ lambda two copy constructor calls
I have the following code snippet. 我有以下代码片段。
#include <iostream>
#include <functional>
using namespace std;
struct A
{
A() { cout << "A "; data = 1; }
A(const A& a) { cout << "cA "; data = a.data; }
~A() { cout << " dA"; }
int data;
};
void f(A& a, function<void(A)> f)
{
cout << "(";
f(a);
cout << ")";
}
int main()
{
A temp;
auto fun = [](A a) {cout << a.data;};
f(temp, fun);
}
The output is: 输出是:
A (cA cA 1 dA dA) dA
A(cA cA 1 dA dA)dA
Why is temp
copied twice? 为什么
temp
复制两次?
I am using Visual C++ (vc140). 我使用的是Visual C ++(vc140)。
function<void(A)>
has a function-call operator with this signature: operator()(A)
ie it takes its argument by value, so calling f(a)
makes a copy. function<void(A)>
有一个带有此签名的函数调用运算符: operator()(A)
即它按值获取其参数,因此调用f(a)
进行复制。
The lambda also takes its argument by value, so when that is called inside the function<void(A)>
call operator another copy gets made. lambda也通过值获取其参数,因此当在
function<void(A)>
call function<void(A)>
内部调用它时,会生成另一个副本。
If you define a move constructor for A
you should see that initializing the lambda argument (from the first copy made by the function
) can be a move instead of a copy, but only if the type has a move constructor. 如果为
A
定义移动构造函数,则应该看到初始化lambda参数(来自function
的第一个副本)可以是移动而不是副本,但仅当类型具有移动构造函数时才可以。 Otherwise it has to be copied. 否则必须复制。
Alternatively, if you use std::function<void(const A&)>
then the call operator will take its argument by reference not by value, so there is only one copy made, to initialize the argument of the lambda. 或者,如果使用
std::function<void(const A&)>
则调用操作符将通过引用而不是值来获取其参数,因此只有一个副本用于初始化lambda的参数。
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