Why can't members and methods have the same name?

[cppaddict]

Hi,

The following class containing a member and a method with the same
name will not compile:

class Test {
private:
bool x;
public:
Test() : x(true) {};
bool x() const {return x;}
};

I am not advocating writing code like the above, but I was puzzled by
the compiler's reason for rejecting it: It said there were multiple
declarations of Test::x.

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

thanks for any explanations,
cpp

 

[JKop]

cppaddict posted:

Hi,

The following class containing a member and a method with the same
name will not compile:

class Test {
private:
bool x;
public:
Test() : x(true) {};
bool x() const {return x;}
};

I am not advocating writing code like the above, but I was puzzled by
the compiler's reason for rejecting it: It said there were multiple
declarations of Test::x.

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

thanks for any explanations,
cpp

int morang(int prot);

int main(void)
{
int (*pMorang)(int) = morang;
}



When you write a function's name without parenthesis, you've got a pointer
to the function.


-JKop

 

[Victor Bazarov]

The following class containing a member and a method with the same
name will not compile:

class Test {
private:
bool x;
public:
Test() : x(true) {};
bool x() const {return x;}
};

I am not advocating writing code like the above, but I was puzzled by
the compiler's reason for rejecting it: It said there were multiple
declarations of Test::x.

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

Are they? Take a look:

struct Functor {
void operator() { 42; }
};

struct Test {
Functor x;
void x() {}
};

int main() {
Test t;
t.x();
}

Now, when I do 't.x()', is that invoking the member _function_ x,
or is it calling the operator() of the _data_ member x?

V

 

[Martin Dickopp]

The following class containing a member and a method with the same
name will not compile:

class Test {
private:
bool x;
public:
Test() : x(true) {};
bool x() const {return x;}
};

I am not advocating writing code like the above, but I was puzzled by
the compiler's reason for rejecting it: It said there were multiple
declarations of Test::x.

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

Are you sure they can always be distiguished syntactically? Please
consider the example below. What should its output be?


#include <iostream>

int foo ()
{
return 1;
}

class Test {
public:
Test () : x (foo) {}
int (*x) ();
int x () const { return 2; }
};

int main ()
{
Test t;
std::cout << t.x () << '\n';
}


Martin

 

[cppaddict]

When you write a function's name without parenthesis, you've got a pointer
to the function.


-JKop

Ah yes. Of course.

Thanks,
cpp

 

[Bill Seurer]

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

&Test::x

which x is it?

 

[=?ISO-8859-15?Q?Juli=E1n?= Albo]

The following class containing a member and a method with the same
name will not compile:

class Test {
private:
bool x;
public:
Test() : x(true) {};
bool x() const {return x;}
};

I am not advocating writing code like the above, but I was puzzled by
the compiler's reason for rejecting it: It said there were multiple
declarations of Test::x.

Since methods and members can always be distinguished syntactically
when they are used, it seems strange that they cannot have the same
name. Just curious why this is?

They can't always be distinguished.

class F {
public:
void operator () () { }
};

class Test {
F f;
void f ();
public:
Test ()
{
f (); // What f ?
}
};