function signatures const vs. non const

[arun]

Hello team,

C++ allows declarations of the fillowing form inside a class, which is
very confusing

void foo();
void foo() const;

My question is when will the second foo() be called as against first
foo()

Similarly

int bar();

const int bar();

Which function will be executed as a result of a call to bar() and
under what circumstances.

Thanks.

[Ian Collins]

Re: function signatures const vs. non const

arun wrote:[color=blue]
> Hello team,
>
> C++ allows declarations of the fillowing form inside a class, which is
> very confusing
>
> void foo();
> void foo() const;
>
> My question is when will the second foo() be called as against first
> foo()
>[/color]
The const form will be called if the object is const, the non-const
otherwise.

--
Ian Collins.

[Ian Collins]

Re: function signatures const vs. non const

arun wrote:[color=blue]
> Similarly
>
> int bar();
>
> const int bar();
>
> Which function will be executed as a result of a call to bar() and
> under what circumstances.
>[/color]
I forgot this bit, you can't overload a const and non-const return with
the same function signature.

--
Ian Collins.

[Rolf Magnus]

Re: function signatures const vs. non const

arun wrote:
[color=blue]
> Hello team,
>
> C++ allows declarations of the fillowing form inside a class, which is
> very confusing[/color]

Only if you don't know what the 'const' means in this place.
[color=blue]
> void foo();
> void foo() const;
>
> My question is when will the second foo() be called as against first
> foo()[/color]

The second will be called if the object it is called for is const or
accessed through a reference or pointer to const. Otheriwse, the first one
is chosen.
[color=blue]
> Similarly
>
> int bar();
>
> const int bar();
>
> Which function will be executed as a result of a call to bar() and
> under what circumstances.[/color]

This shouldn't compile, because the functions have the same signature.

[benben]

Re: function signatures const vs. non const

arun wrote:[color=blue]
> Hello team,
>
> C++ allows declarations of the fillowing form inside a class, which is
> very confusing
>
> void foo();
> void foo() const;
>
> My question is when will the second foo() be called as against first
> foo()[/color]

class Circle
{
double radius;
public:
Circle(double r):radius(r){}

double& r(){return radius;}
const double& r() const{return radius;}
};

int main()
{
Circle c1(10);
const Circle c2(20);

// You should be able to read
// a const circle's radius:

double r1 = c1.r(); // calls Circle::r()
double r2 = c2.r(); // calls Circle::r() const

// But not to write to:

c1.r() = 3.5; // OK, writing to a double&
c2.r() = 3.5; // ERROR, cannot write to a const double&
}
[color=blue]
>
> Similarly
>
> int bar();
>
> const int bar();
>
> Which function will be executed as a result of a call to bar() and
> under what circumstances.[/color]

Under no circumstances because this won't compile.
[color=blue]
>
> Thanks.
>[/color]

Regards,
Ben