overloading operator=

[Daniel Allex]

I have the following code:

#include "iostream.h "

class a
{
public:
struct my_struct
{
int one;
int two;
};

my_struct operator= (int);
};


my_struct a::operator=(in t op)
{

return 1;
}

main()
{
cout<<"Test"<<e ndl;
return 0;
}

I get the following compiler errors:

C:\Code\test\te st.cpp(16) : error C2143: syntax error : missing ';'
before 'tag::id'
C:\Code\test\te st.cpp(16) : error C2501: 'my_struct' : missing
storage-class or type specifiers
C:\Code\test\te st.cpp(16) : fatal error C1004: unexpected end of file
found

If I change the return type to and int or some other defined type for
the overloaded operator, it compiles fine.

[Josephine Schafer]

Re: overloading operator=


"Daniel Allex" <dallex@erols.c om> wrote in message
news:3F4D735D.2 14BFD54@erols.c om...[color=blue]
> I have the following code:
>
> #include "iostream.h "[/color]
#include <iostream>
[color=blue]
>
> class a
> {
> public:
> struct my_struct
> {
> int one;
> int two;
> };
>
> my_struct operator= (int);[/color]

The return type of assignment operator is the left hand side operand (object
on which the
function is invoked). So my_struct as return type is incorrect. Also one
should return by reference.
So it's return type should be a& (reference to a).
a& operator= (int);
[color=blue]
> };
>
>
> my_struct a::operator=(in t op)[/color]
Same comment as above.[color=blue]
> {
>
> return 1;[/color]

return *this;[color=blue]
> }
>[/color]
[color=blue]
> main()[/color]

main always returns int.
int main ()[color=blue]
> {
> cout<<"Test"<<e ndl;[/color]

Everything in standard headers is now in std namespace.
So above line should be -
std::cout << "Test"<<std::en dl;
[color=blue]
> return 0;
> }
>[/color]

HTH.

--
J.Schafer