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c++functionoperator-overloadingcout
Engineer999
Engineer999

Reputation: 3955

Overloading input / output operators

When overloading the input or output operators (>> or <<) , I see that the function takes two arguments, a reference to an iostream object and a reference to whatever object we wish to input or output. An operator<< example for a Matrix object :

std::ostream& operator<< (std::ostream& stream, const Matrix& matrix);

When calling this function we only need to pass the Matrix object, for example :

Matrix m1;
cout<<m1;

What about the other ostream object argument? How does this work? Thanks in advance

Upvotes: 1

Views: 664

Answers (3)

utnapistim
utnapistim

Reputation: 27365

What about the other ostream object argument? How does this work?

If the compiler finds:

a = b <op> c;

(where <op> can be +, -, ==, >>, etc), the compiler will try the following resolutions:

a = operator <op>(b, c); // (1)
a = b.operator <op> (c); // (2)

This depends on the type of operator (for example, + is a binary operator and cannot be implemented in form (2)) and implementations available (the code you write and the resolution priority).

For operator << (your example above), these pieces of code are equivalent:

1:

Matrix m1;
cout<<m1;

2:

Matrix m1;
operator<<(cout, m1);

For native data types, std:::istream offers a member implementation:

int i;
cout.operator<<(i);

Upvotes: 1

SumitV
SumitV

Reputation: 306

Consider this case:

cout<<m1

Here cout is the first argument, m1 is second. The return value of this method call is not captured by your code.

Case 2:

cout << m1 <<  m2;

Here there will be two calls to operator. Return of the first call is passed as argument of next call.

Upvotes: 0

Puppy
Puppy

Reputation: 146940

The other argument is cout itself. The call becomes operator<<(cout, m1).

If the operator only took one argument, you could write << m1; as a complete statement, which you cannot.

Upvotes: 5

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