Question

C++’s std::cout seems to be an interesting thing. I tried the following program on my C++ compiler today:

cout<<"!"<<"@"<<endl;
cout<<"!"<<cout<<"@"<<endl;
cout<<"!"<<(cout<<"@")<<endl;

And the outputs are rather curious:

!@
!0x601068@
@!0x601068

The first line is pedestrian; the second is understandable; however, the third line is beyond my knowledge. Could someone explain the output? Thank you guys all in advance!

Ziyao Wei

Answer 1

This line:

cout<<"!"<<(cout<<"@")<<endl;

It is first executing:

(cout << "@")

Note: It could have executed something else first, but the compiler optimized this sub-expression and found it could move this to the start of the statement without breaking any constraints.

The result of this expression is a stream (cout). So the resulting expression is:

cout<<"!"<< cout <<endl;

This results in:

@!<pointer>