C++ operator overloading error


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# 1  
C++ operator overloading error

hi,
I defined my own String class and overloaded the output operator with
Code:
friend ostream& operator<<(ostream& os, const myString& str);
//overloads the << operator so it can be used to output values of type myString

which works fine. Until I try to execute the following statement:
Code:
// + concatenation operator is already defined for myString
cout << (str4=str2+str) << endl; //This statement concatenates str to str2, then assigns it to str4 and returns str4, right?

And, I get error
Code:
error: no match for ‘operator<<’ (operand types are ‘std::ostream’ {aka ‘std::basic_ostream<char>’} and ‘void’)
  cout << (str4=str2+str) << endl;
  ~~~~~^~~~~~~~~~~~~~~~~~

The above code works fine when I change the concatenation statement as follows:
Code:
str4=str2+str;
cout << str4 << endl;

What did I miss?
# 2  
What type does your 'operator =' return?
# 3  
Quote:
Originally Posted by Corona688
What type does your 'operator =' return?
Code:
void operator=(const myString& rside);

# 4  
That means the expression 'str2=whatever' returns void -- or rather, returns nothing. So there's nothing to cout.

Instead have it return const myString & and at the end, put return(*this);
# 5  
I changed the member function like this:
Code:
myString operator=(const myString& rside)

and return rside at the end.
# 6  
Returning rside is clever and ought to work, but return-by-value is a bad idea -- that makes a local copy, which isn't just wasteful, in some circumstances that's an infinite recursion and out-of-memory crash. Assignment operators shouldn't use copy constructors.

You pretty much have to return a reference here.

Code:
const myString &operator=(const myString& rside)

This User Gave Thanks to Corona688 For This Post:
# 7  
makes sense. Thanks for the help.
This User Gave Thanks to milhan For This Post:
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