Ostream& operator<< overloading code not working

Question

#include <string> #include <iostream> template <typename T> T max(T a, T b) { return a > b ? a : b; } class Dummy { private: std::string name; int age; public: Dummy(int an_age) {age = an_age;} bool operator> (Dummy &a) {return age > a.age;} std::string toString() const {return "The age is " + age;} }; std::ostream& operator<<(std::ostream& out, const Dummy& d) {return out<< d.toString();} int main() { std::cout << max(3, 7) << std::endl; std::cout << max(3.0, 7.0) << std::endl; std::cout << max<int>(3, 7.0) << std::endl; std::cout << max("hello", "hi") << std::endl; Dummy d1(10); Dummy d2(20); std::cout << max(&d1, &d2) << std::endl; return 0; }

I'm pretty new to C++ but not new to programming. I've written the code to play with template and operator overloading in C++.

It took quite a while to make it compile and partially work.

The ostream operator<< is not working properly, only to return the address of the object. I can't figure out the causes.
I managed to make it compile by blind trial and error, so I suspect the code might be broken to some extent. And I may not be aware of what'd be improved.

Nikolai Fetissov · Accepted Answer · 2011-11-22 20:54:30Z

6

Your max(&d1,&d2) expression gives you the address, and that is printed out. Your operator overloading is fine.

answered Nov 22, 2011 at 20:54

Nikolai Fetissov

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7 Comments

Tim Over a year ago

also, "The age is " + age will NOT do what you want. You want to use std::ostringstream.

Nikolai Fetissov Over a year ago

Yes, max("hello", "hi") is wrong too. OP should read up on C++ built-in types.

Seth Carnegie Over a year ago

@Tim or just string::operator+ for this simple case

Tim Over a year ago

@Seth, std::string does not have an operator+(int)

Walter Over a year ago

std::string::operator+(int) is not called in here at all. What is called is operator+(const char*,int) followed by std::string(const char*).

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Michael · Accepted Answer · 2011-11-22 20:56:49Z

I assume the line you're talking about is

std::cout << max(&d1, &d2) << std::endl;

The problem is you are passing Dummy * instead of Dummy. That makes max return Dummy *, and since your overloaded operator<< takes (essentially) Dummy, it isn't invoked. If you're trying to pass by reference, you don't need to do anything special on the caller side, just make the function take a reference and the compiler will figure it out.

Yakov Galka · Accepted Answer · 2011-11-22 21:00:47Z

Don't write your own max, use the standard one instead:
```
#include <algorithm> void f() { int a = std::max(8, 4); } 
```
The only difference is that the standard max uses operator < by default, just like everything else in the standard library.
Your toString function does something different from what you think it does. It instead returns the sub string of "The age is " starting at the character number age. For example if age is 3, toString will return " age is ". To convert the integer to string you have to use ostringstream:
```
std::string toString() const { std::ostringstream s; s << "The age is " << age; return s.str(); } 
```

std::max is avoiding unnecessary copies as well (passing by reference vs copying)

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