Function that prints something to std::ostream and returns std::ostream?

Question

I want to write a function that outputs something to a ostream that's passed in, and return the stream, like this:

std::ostream& MyPrint(int val, std::ostream* out) {
  *out << val;
  return *out;
}

int main(int argc, char** argv){
    std::cout << "Value: " << MyPrint(12, &std::cout) << std::endl;
    return 0;
}

It would be convenient to print the value like this and embed the function call in the output operator chain, like I did in main().

It doesn't work, however, and prints this:

$ ./a.out
12Value: 0x6013a8

The desired output would be this:

Value: 12

How can I fix this? Do I have to define an operator<< instead?

UPDATE: Clarified what the desired output would be.

UPDATE2: Some people didn't understand why I would print a number like that, using a function instead of printing it directly. This is a simplified example, and in reality the function prints a complex object rather than an int.

Answer 1

Since C++11, chaining a function returning a reference to an ostream with another ostream is ill-formed. For e.g., see this answer.

The reason is because since C++11, there is no implicit conversion of an ostream to a void *.

Answer 2

The problem can be solved this way:

Answer 3

You can't fix the function. Nothing in the spec requires a compiler to evaluate a function call in an expression in any particular order with respect to some unrelated operator in the same expression. So without changing the calling code, you can't make MyPrint() evaluate after std::cout << "Value: "

Left-to-right order is mandated for expressions consisting of multiple consecutive << operators, so that will work. The point of operator<< returning the stream is that when operators are chained, the LHS of each one is supplied by the evaluation of the operator to its left.

You can't achieve the same thing with free function calls because they don't have a LHS. MyPrint() returns an object equal to std::cout, and so does std::cout << "Value: ", so you're effectively doing std::cout << std::cout, which is printing that hex value.

Since the desired output is:

Value: 12

the "right" thing to do is indeed to override operator<<. This frequently means you need to either make it a friend, or do this:

class WhateverItIsYouReallyWantToPrint {
    public:
    void print(ostream &out) const {
        // do whatever
    }
};

ostream &operator<<(ostream &out, const WhateverItIsYouReallyWantToPrint &obj) {
    obj.print(out);
}

If overriding operator<< for your class isn't appropriate, for example because there are multiple formats that you might want to print, and you want to write a different function for each one, then you should either give up on the idea of operator chaining and just call the function, or else write multiple classes that take your object as a constructor parameter, each with different operator overloads.

Answer 4

After changing the pointer to a reference, you can do this:

#include <iostream>

std::ostream& MyPrint(int val, std::ostream& out) {
    out << val;
    return out;
}

int main(int, char**) {
    MyPrint(11, std::cout << "Value: ") << std::endl; 

    return 0;
}

The syntax for MyPrint is essentially that of an unrolled operator<< but with an extra argument.

Answer 5

You have two options. The first, using what you already have is:

std::cout << "Value: ";
MyPrint(12, &std::cout);
std::cout << std::endl;

The other, which is more C++-like, is to replace MyPrint() with the appropriate std::ostream& operator<<. There's already one for int, so I'll do one just a tad more complex:

#include <iostream>

struct X {
    int y;
};

// I'm not bothering passing X as a reference, because it's a
// small object
std::ostream& operator<<(std::ostream& os, const X x)
{
    return os << x.y;
}

int main()
{
    X x;
    x.y = 5;
    std::cout << x << std::endl;
}

Answer 6

There's no way to do what you're expecting there because of the order the functions are evaluated in.

Is there any particular reason you need to write directly to the ostream like that? If not, just have MyPrint return a string. If you want to use a stream inside MyPrint to generate the output, just use a strstream and return the result.

Answer 7

You want to make MyPrint a class with friend operator<<:

class MyPrint
{
public:
    MyPrint(int val) : val_(val) {}
    friend std::ostream& operator<<(std::ostream& os, const MyPrint& mp) 
    {
        os << mp.val_;
        return os;
    }
private:
    int val_;
};

int main(int argc, char** argv)
{
    std::cout << "Value: " << MyPrint(12) << std::endl;
    return 0;
}

This method requires you to insert the MyPrint object into the stream of your choice. If you REALLY need the ability to change which stream is active, you can do this:

class MyPrint
{
public:
    MyPrint(int val, std::ostream& os) : val_(val), os_(os) {}
    friend std::ostream& operator<<(std::ostream& dummy, const MyPrint& mp) 
    {
        mp.os_ << mp.val_;
        return os_;
    }
private:
    int val_;
    std::ostream& os_
};

int main(int argc, char** argv)
{
    std::cout << "Value: " << MyPrint(12, std::cout) << std::endl;
    return 0;
}

Answer 8

In your case the answer is obviously:

 std::cout << "Value: " << 12 << std::endl;

If that isn't good enough, please explain what output you want to see.

Answer 9

First, there is no reason not to pass in the ostream by reference rather than by a pointer:

std::ostream& MyPrint(int val, std::ostream& out) {
  out << val;
  return out;
}

If you really don't want to use std::ostream& operator<<(std::ostream& os, TYPE), you can do this:

int main(int argc, char** argv){
    std::cout << "Value: ";
    MyPrint(12, std::cout) << std::endl;
    return 0;
}