c++11 为什么使用ref，和引用的区别

Posted 2022-06-20 guxuanqing

std::ref只是尝试模拟引用传递，并不能真正变成引用，在非模板情况下，std::ref根本没法实现引用传递，只有模板自动推导类型时，ref能用包装类型reference_wrapper来代替原本会被识别的值类型，而reference_wrapper能隐式转换为被引用的值的引用类型。

std::ref主要是考虑函数式编程（如std::bind）在使用时，是对参数直接拷贝，而不是引用

其中代表的例子是thread
比如thread的方法传递引用的时候，必须外层用ref来进行引用传递，否则就是浅拷贝。

线程函数的参数按值移动或复制。如果引用参数需要传递给线程函数，它必须被包装（例如使用std :: ref或std :: cref）。

#include <functional>
#include <iostream>
#include<cstring>
#include<string.h>
#include<memory>
#include<atomic>
#include<unordered_map>
#include<mutex>
#include <thread>
#include <string>
void method(int & a) a += 5;

using namespace std;
int main()

    int a = 0;
    // each reference used by the threads would refer to the same object.
    thread th(method,ref(a));
    th.join();
    cout << a <<endl;
    /*
     *
I could compile your code successfully with MSVC2013. However, thread() works passing copies of its argument to the new thread. This means that if your code would compile on your compiler, each thread wourd run with its own copy of ht, so that at the end, main‘s a would be empty.
GCC doesn‘t compile with this weird message.
     */
    /*thread th2(method, a);  //浅拷贝
    th2.join();
    cout << a <<endl;*/
    return 0;

 

/** @file  bindRefT.cpp
*  @note   
*  @brief
*  @author 
*  @date   2019-8-8
*  @note   
*  @history
*  @warning
*/
#include <functional>
#include <iostream>
//std::ref主要是考虑函数式编程（如std::bind）在使用时，是对参数直接拷贝，而不是引用
void f(int& n1, int& n2, const int& n3)

    std::cout << "In function: " << n1 << ‘ ‘ << n2 << ‘ ‘ << n3 << ‘\n‘;
    ++n1; // increments the copy of n1 stored in the function object
    ++n2; // increments the main()‘s n2
    // ++n3; // compile error


int main()

    int n1 = 1, n2 = 2, n3 = 3;
    std::function<void()> bound_f = std::bind(f, n1, std::ref(n2), std::cref(n3));
    n1 = 10;
    n2 = 11;
    n3 = 12;
    std::cout << "Before function: " << n1 << ‘ ‘ << n2 << ‘ ‘ << n3 << ‘\n‘;
    bound_f();
    std::cout << "After function: " << n1 << ‘ ‘ << n2 << ‘ ‘ << n3 << ‘\n‘;