default arg value lost via function pointer

[Metaosp]

Hi,

How can I call a function through function pointer while keeping the
default argument value working? I mean:

void foo(int i=3) {
cout << i << endl;
}

void (*bar)(int);

int main() {
bar = &foo;

foo(); // ok, prints '3'
bar(1); // ok, prints '1'

bar(); // doesn't compile...
// default argument lost via function pointer
}



Thanks,
Metaosp

 

[Sunil Varma]

Hi,

How can I call a function through function pointer while keeping the
default argument value working? I mean:

void foo(int i=3) {
cout << i << endl;
}

void (*bar)(int);

int main() {
bar = &foo;

foo(); // ok, prints '3'
bar(1); // ok, prints '1'

bar(); // doesn't compile...
// default argument lost via function pointer
}



Thanks,
Metaosp

Please change your declaration of function pointer to

void (*bar)(int = 3);

now you can call the bar with or without pointer.

As per the declation of the bar function pointer, the compiler assumes
that the pointer will be called with one int argument.

The compiler doesnt what will be stored in bar, whose value is
asssigned runtime.

Regards
Sunil Varma

 

[Metaosp]

Please change your declaration of function pointer to

void (*bar)(int = 3);

now you can call the bar with or without pointer.

I've tried this, but it doesn't compile... the error message I got is
"default arguments are only permitted for function parameters". Any
idea?


Thanks,
Metaosp

 

[Sunil Varma]

May I know the compiler you are using?

I tried in gcc 3.4.2 and Borand C++ 5.5.1.

It worked fine.

Regards
Sunil Varma

 

[Joe Van Dyk]

Hi,

How can I call a function through function pointer while keeping the
default argument value working? I mean:

void foo(int i=3) {
cout << i << endl;
}

void (*bar)(int);

int main() {
bar = &foo;

foo(); // ok, prints '3'
bar(1); // ok, prints '1'

bar(); // doesn't compile...
// default argument lost via function pointer
}

I don't believe you can. Default arguments aren't considered part of a
function's type.

I'm suprised this isn't a FAQ -- I can't find it at
http://www.parashift.com/c++-faq-lite.

Joe

 

[Metaosp]

May I know the compiler you are using?

I tried in gcc 3.4.2 and Borand C++ 5.5.1.

It worked fine.

Interesting, I am using gcc 4.0.2. I wonder whether this is an
undefined behavior of C++ or a bug in gcc 4?


Metaosp

 

[Rolf Magnus]

Interesting, I am using gcc 4.0.2. I wonder whether this is an
undefined behavior of C++ or a bug in gcc 4?

It's a bug in gcc 3.4.2 and Borland C++ 5.5.1.

 

[Metaosp]

It's a bug in gcc 3.4.2 and Borland C++ 5.5.1.

So this is a defined (or induced) limitation of C++ function pointers?


Metaosp

 

[Jonathan Mcdougall]

How can I call a function through function pointer while keeping the
default argument value working? I mean:

This is explicitly forbidden by the standard :

"This means that default arguments cannot appear, for example, in
declarations of pointers to functions, references to functions, or
typedef declarations." (8.3.6§3 note 88)

void foo(int i=3) {
cout << i << endl;
}

void (*bar)(int);

int main() {
bar = &foo;

foo(); // ok, prints '3'
bar(1); // ok, prints '1'

bar(); // doesn't compile...
// default argument lost via function pointer
}

Jonathan

 

[Rolf Magnus]

So this is a defined (or induced) limitation of C++ function pointers?

Yes.