Called a function with "cout" statement inside a "cout" statement

Question

I came across this rather vague behavior when messing around with code , here's the example :

#include <iostream>

using namespace std;


int print(void);

int main(void)
{
    cout << "The Lucky " << print() << endl;     //This line
    return 0;
}

int print(void)
{
    cout << "No : ";
    return 3;
}

In my code, the statement with comment //This lineis supposed to print out The Lucky No : 3, but instead it was printed No : The Lucky 3. What causes this behavior? Does this have to do with C++ standard or its behavior vary from one compiler to another?

Helaina

Why do you believe that your expectation instead of the actual outcome is the more logical one? Can you explain your reasoning? – Eneidaenema

Sidenote: Your print function is a nice example of a function with side-effects. Besides calculating a return value, it also changes global program state. Such functions are usually more difficult to reason about. Generally, strive towards functions that either compute a value or change state, but not both simultaneously. – Eneidaenema

Answer 1

The order of evaluation of arguments to a function is unspecified. Your line looks like this to the compiler:

operator<<(operator<<(operator<<(cout, "The Lucky "), print()), endl);

The primary call in the statement is the one with endl as an argument. It is unspecified whether the second argument, endl, is evaluated first or the larger sub-expression:

operator<<(operator<<(cout, "The Lucky "), print())

And breaking that one down, it is unspecified whether the function print() is called first, or the sub-expression:

operator<<(cout, "The Lucky ")

So, to answer your question:

What causes this behavior? Does this has to do with C++ standard or its behavior vary from one compiler to another?

It could vary from compiler to compiler.

Execrate

That's totally possible, and I might begin to see it ;) – Plum

@atamanroman: In the following expression: operator<<(operator<<(cout, "The Lucky "), print()) What do you think should happen first, This: operator<<(cout, "The Lucky "), or this: print()? Justify your answer. – Execrate

I know, It's not defined and I get it know. So cout << "The lucky " + print() would be well defined? Deleted my answer (+1). – Plum

@atamanroman: Just for the sake of preciseness, you might want to be aware that there's a difference between "undefined" and "unspecified" behaviour. Roughly speaking, undefined behaviour means that anything can happen; unspecified behaviour means that what happens depends on the implementation (but implementations are not required to document what they do). – Thaumatrope

@atamanroman: the behavior isn't undefined, it's unspecified -- there's a difference. The compiler is free to evaluate the parameters in whatever order (including a random order) that it finds best, it's not free to not evaluate a parameter at all or to blow up your toaster. – Chanel

@atamanroman: cout << "The lucky" + print() would output No : lucky. – Perpetuity

@BenjaminLindley Can you please also explain how the statments translated to 'operator<<(operator<<(operator<<(cout, "The Lucky "), print()), endl); ' – Expressman

Answer 2

Let's call the operator << simply operator .

Now we can write

cout << "The Lucky"

as

operator(cout, "The Lucky")

The result of this operation is cout, and it is passed to next <<, so we can write

operator(operator(cout, "The Lucky"), print() )

It is a function invocation with two parameters, and the standard doesn't say anything about the order of their evaluation.

So with some compilers you really may get

The Lucky No : 3

Dancette

Answer 3

In my compiler No: The Lucky 3 is the output.... it means that its behaviour varies from compiler to compiler.

Bedel