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:perator=(int op)
{

return 1;
}

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

I get the following compiler errors:

C:\Code\test\test.cpp(16) : error C2143: syntax error : missing ';'
before 'tag::id'
C:\Code\test\test.cpp(16) : error C2501: 'my_struct' : missing
storage-class or type specifiers
C:\Code\test\test.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]

I have the following code:

#include "iostream.h"

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

my_struct operator= (int);

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);

};


my_struct a:perator=(int op) Same comment as above.
{

return 1;

return *this;

}

main()

main always returns int.
int main ()

{
cout<<"Test"<<endl;

Everything in standard headers is now in std namespace.
So above line should be -
std::cout << "Test"<<std::endl;

return 0;
}

HTH.