ostream chaining, output order

Question

I have a function that takes an ostream reference as an argument, writes some data to the stream, and then returns a reference to that same stream, like so:

Answer 1

In your statement std::cout << "Hello, world!" << print( std::cout ) << std::endl it's undefined whether std::cout << "Hello, world!" happens before or after print( std::cout ). That's why the order may not be what you expect.

The hex value comes from the fact that you're also doing std::cout << std::cout (print returns std::cout which is fed into the << chain). The right hand std::cout is converted to a void * and that's printed to the output.

Answer 2

The reason is that your print() function will be evaluated before the rest of the statement and return a reference to cout which is then actually printed as a pointer (cout << cout). This order of evaluation is actually unspecified behavior, but seems to be the case with your compiler.

As for defining a stream aware "function" that actually has defined behavior with the same functionality, this would work;

#include <iostream>

template <class charT, class traits>
  std::basic_ostream<charT,traits>& print ( std::basic_ostream<charT,traits>& os )
{
        os << " How are you?" << std::endl;
        return os;
}

int main() {
  std::cout << "Hello, world!" << print << std::endl;
}

See also this answer for a little more detail on what "unspecified" actually means in this case.

Answer 3

The behavior of your code is unspecified as per the C++ Standard.

Explanation

The following (I removed std::endl for simplicity)

std::cout << "Hello, world!" << print( std::cout ); 

is equivalent to this:

operator<<(operator<<(std::cout, "Hello, World!"), print(std::cout));

which is a function call, passing two arguments:

• First argument is : operator<<(std::cout, "Hello, World!")
• Second argument is : print(std::cout)

Now, the Standard doesn't specify the order in which arguments are evaluated. It is unspecified. But your compiler seems to evaluate the second argument first, that is why it prints "How are you?" first, evaluating the second argument to a value of type std::ostream& which then gets passed to the call shown above (that value is the object std::cout itself).

Why hexadecimal output?

You get hexadecimal output because the second argument evaluates to std::cout, which is being printed as hexadecimal number, because std::cout implicitly converts into pointer value of void* type, which is why it is printed as hexadecimal number.

Try this:

void const *pointer = std::cout; //implicitly converts into pointer type!
std::cout << std::cout << std::endl;
std::cout << pointer << std::endl;

It will print the same value for both. For example, this example at ideone prints this:

0x804a044
0x804a044 

Also note that I didn't use explicit cast; rather std::cout is implicitly converted into pointer type.

Hope that helps.

---

What is the proper way to write a function that inserts data into an ostream but that can also chain with operator<<?

When it depends on what you mean by chaining? Obviously, the following wouldn't work (as explained above):

std::cout << X << print(std::cout) << Y << Z; //unspecified behaviour!

No matter how you write print().

However this is well-defined:

print(std::cout) << X << Y << Z; //well-defined behaviour!

Answer 4

This would work, to combine print with << and control the order:

print( std::cout << "Hello, world!" ) << std::endl;

Or, if you want a function that's called with <<, see Joachim's answer.

Answer 5

Hexadecimal Output

Before C++11, the class std::ostream has a conversion function to void*. Since your print function returns std::ostream&, when evaluating std::cout << print(...), the returned std::ostream lvalue will be implicitly converted to void* and then be outputted as a pointer value. This is why there is a hexadecimal output.

Since C++11, this conversion function is replaced by an explicit conversion function to bool, so trying to output an std::ostream object becomes ill-formed.

Evaluation Order

Before C++17, overloaded operator is considered a function call for analyzing evaluation order, and evaluation order of different arguments of a function call is unspecified. So it is not strange that the print function is evaluated firstly, which causes How are you? is outputted firstly.

Since C++17, the evaluation order of operands of operator << is strictly from left to right, and operands of overloaded operator share the same evaluation order as those of the bulit-in one (see more details here). So your program will always get the output (assume print returns something able to be outputted)

Hello, world! How are you?
something returned by print

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