Re: The order of calling functions and << <<

[Alex Buell <alex dot buell at tahallah dot demon dot co dot uk>]

On Mon, 7 Dec 1998, Joe Buck wrote:

> The standard says that the behavior of the above program is undefined,
> because the order of evaluation of arguments to a function call is unspecified.
> 
> You have written
> 
> operator<<(operator<<(operator<<(operator<<(cout, "1"),foo()),"4"),endl);
> 
> The source of the undefined behavior is that in
> 
> operator<<(operator<<(cout, "1"), foo());
> 
> the compiler is free to either do operator<<(cout,"1") first or to call
> foo() first.  This means that you could get "124\n" or "214\n".

Um! I don't think that is right either. The foo() function prints 2 *and*
returns an integer - 3 in this case. So I believe what would happen with
the line:

>   cout << "1" << foo() << "4" << endl;

would become

1234

This makes sense. Or not? Hmm. Lemme try this out. 

Cheers,
Alex.

---
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