为什么分配给std :: ostream的引用无法编译？

Question

我正在使用抽象类std :: ostream.有以下参考:

std::ostream &o = std::cout; 

如果满足任何条件,我需要初始化o,以便输出重定向到std :: cout.如果没有,输出将被重定向到文件

if (!condition)
    o = file; //Not possible 

如何正确编写代码？

Answer 1

或者:

std::ostream &o = condition ? std::cout : file;

或者如果两个代码段之间有代码:

std::ostream *op = &std::cout;
// more code here
if (!condition) {
    op = &file;
}
std::ostream &o = *op;

问题不是与抽象类有关,而是引用无法重新定位.

表达式的含义o = file不是"make orefer to file",而是"将值复制file到referand中o".幸运的是,std::ostream没有operator=,所以它无法编译,std::cout也没有修改.但考虑一下不同类型的情况:

#include <iostream>

int global_i = 0;

int main() {
    int &o = global_i;
    int file = 1;
    o = file;
    std::cout << global_i << "\n"; // prints 1, global_i has changed
    file = 2;
    std::cout << o << "\n"; // prints 1, o still refers to global_i, not file
}

Answer 2

您无法重置参考.

这是另一种选择:

std::ostream &o = (!condition) ? file : std::cout;