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?

Why do you believe that your expectation instead of the actual outcome is the more logical one? Can you explain your reasoning? — stakx - no longer contributing, Jul 22 '12 at 20:55
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. — stakx - no longer contributing, Jul 22 '12 at 21:04

score 18 · Accepted Answer · answered Jul 22 '12 at 21:01

18

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.

answered Jul 22 '12 at 21:01

Benjamin Lindley

101,917
9
204
274

That's totally possible, and I might begin to see it ;) – atamanroman Jul 22 '12 at 21:07
@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. – Benjamin Lindley Jul 22 '12 at 21:10
I know, It's not defined and I get it know. So cout << "The lucky " + print() would be well defined? Deleted my answer (+1). – atamanroman Jul 22 '12 at 21:14
1

@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). – Stuart Golodetz Jul 22 '12 at 21:31
1

@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. – jmoreno Jul 22 '12 at 21:35
1

@atamanroman: `cout << "The lucky" + print()` would output `No : lucky`. – molbdnilo Jul 23 '12 at 12:16
@BenjaminLindley Can you please also explain how the statments translated to 'operator<<(operator<<(operator<<(cout, "The Lucky "), print()), endl); ' – Dexters Oct 03 '13 at 04:32

score 4 · Answer 2 · edited Jul 23 '12 at 01:07

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

score 0 · Answer 3 · answered Nov 20 '15 at 17:11

0

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

answered Nov 20 '15 at 17:11

Azhar Sindhi

37
5

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

3 Answers3