反直觉C++🔥, 👏欢迎github star

引用有时会失效

#include <functional>
#include <iostream>
 
void fun(int& n) {
    std::cout << "In fun " << "n:" << n << '\n';
    ++n;
    std::cout << "In fun " << "n:" << n << '\n';
}
 
int main() {
    int n = 1;
    std::cout << "Before fun " << "n:" << n << '\n';
    fun(n);
    std::cout << "After fun " << "n:" << n << '\n';

    n = 10;
    std::function<void()> bound_fun = std::bind(fun, n);
    n = 100;
    std::cout << "Before bound_fun " << "n:" << n << '\n';
    bound_fun();
    std::cout << "After bound_fun " << "n:" << n << '\n';
    
    return 0;
}

输出

Before fun n:1
In fun n:1
In fun n:2
After fun n:2
Before bound_fun n:100
In fun n:10
In fun n:11
After bound_fun n:100

fun函数的参数n类型为int&，直接调用fun时，左值引用跟随变化，然而当左值引用遇到std::bind时，左值引用却失效了:fun函数在std::bind时n为10;随后n被赋值100，100并没有被传递给bound_fun，bound_fun携带的仍是std::bind时的10，有点反直觉。

奇怪的知识增加了

std::bind有一个特殊点就是参数列表的形参是在定义时就决定的（即使传递的是左值引用），如果想要前后联动需配合std::ref。

std::ref示例

#include <functional>
#include <iostream>
 
void fun(int& n1, int& n2, const int& n3) {
    std::cout << "In function " << "n1:" << n1 << " n2:" << n2 << " n3:" << n3 << '\n';
    ++n1; // 增加存储于函数对象的 n1 副本
    ++n2; // 增加 main() 的 n2
    // ++n3; // 编译错误
}
 
int main() {
    int n1 = 1, n2 = 2, n3 = 3;
    std::function<void()> bound_fun = std::bind(fun, n1, std::ref(n2), std::cref(n3));
    n1 = 10;
    n2 = 11;
    n3 = 12;
    std::cout << "Before function " << "n1:" << n1 << " n2:" << n2 << " n3:" << n3 << '\n';
    bound_fun();
    std::cout << "After function " << "n1:" << n1 << " n2:" << n2 << " n3:" << n3 << '\n';
}

输出

Before function n1:10 n2:11 n3:12
In function n1:1 n2:11 n3:12
After function n1:10 n2:12 n3:12