Why does assigning to a reference to std::ostream fail to compile?

Question

I am using an abstract class std::ostream. There is the following reference:

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

If any condition is met I need to initialize o so that the output will be redirected to std::cout. If not, output will be redirected to the file

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

How to correctly write the code?

Answer 1

Either:

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

or if there's code in between your two snippets:

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

The problem isn't specifically to do with the abstract class, it's that references cannot be reseated.

The meaning of the expression o = file isn't "make o refer to file", it is "copy the value of file into the referand of o". Luckily for you, std::ostream has no operator=, and so it fails to compile and std::cout isn't modified. But consider what happens with a different type:

#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
}