c++11 為什么使用ref，和引用的區別

std::ref只是嘗試模擬引用傳遞，並不能真正變成引用，在非模板情況下，std::ref根本沒法實現引用傳遞，只有模板自動推導類型時，ref能用包裝類型reference_wrapper來代替原本會被識別的值類型，而reference_wrapper能隱式轉換為被引用的值的引用類型。

std::ref主要是考慮函數式編程（如std::bind）在使用時，是對參數直接拷貝，而不是引用

其中代表的例子是thread
比如thread的方法傳遞引用的時候，必須外層用ref來進行引用傳遞，否則就是淺拷貝。

線程函數的參數按值移動或復制。如果引用參數需要傳遞給線程函數，它必須被包裝（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);  //淺拷貝
    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主要是考慮函數式編程（如std::bind）在使用時，是對參數直接拷貝，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';
}