The Assignment operator=
Author
Consider:
class Test
{
public:
Test& operator=( const Test& t )
{
cout << "Assignment\n";
m_Val = t.m_Val;
return( *this );
}
int m_Val;
};
void func( Test t )
{
cout << "func: " << t.m_Val << "\n";
}
void main()
{
Test T1;
T1.m_Val = 5;
Test T2 = T1; //assignment operator is not called here but members are copied
func( T1 ); //assignment operator not called, members are copied
T2 = T1; //assignment operator called
}
Why is it for the case of Test T2 = T1 the assignment operator does not get called? How do we fix this?
Author
Well done!
Interestingly, if you have a function that returns an object and assign it to something else, the copy constructor is called followed by the assignment operator.
Test func()
{
Test t;
return( t );
}
Test t = func(); //only copy constructor
t = func(); //copy constructor then assignment
[Edited by - Simplicity on April 17, 2006 3:42:43 PM]
Interestingly, if you have a function that returns an object and assign it to something else, the copy constructor is called followed by the assignment operator.
Test func()
{
Test t;
return( t );
}
Test t = func(); //only copy constructor
t = func(); //copy constructor then assignment
[Edited by - Simplicity on April 17, 2006 3:42:43 PM]
To expand a bit, although "Text t = func();" sure looks like it's using the assignment operator it really isn't. It's really just syntax-suger for "Text t(func());".
This topic is closed to new replies.
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