Default function parameter values
 

Hi,

how can I set default values to template parameters?

I've this code:

class A
{
template<typename T>
void foo( string = "", T* = 0, bool (T::*func)(void) = 0 );
};

when I invoke A::foo with the appropriate 3 parameters, everything
works fine. However, I'd like to invoke the function foo also
with the first parameter, like objectA.foo( string("test") );
When I try to compile, I get the error message:
"no matching function for A::foo(std::string)".

Why? I set the second and third parameter of foo to default values,
so when the second and third argument are not specified, they should
be set to 0. But this seems not to work that way.

Tim

Re: Default function parameter values
 

> Because the compiler needs to know what 'T' is. It cannot deduce
> the type from an integral expression ('0'). You need to tell it
> what the type T is by supplying the template argument, like
>
> objectA.foo<sometype>(string("test"));

But why can I invoke the function with:

objectA.foo( string( "test" ), objectA, &A::funcPtr );

where I do not specify the type of T either?
Where does the compiler get to information what type T is?

Re: Default function parameter values
 

Victor Bazarov wrote:
> ...
> Set to 0 of what type? '0' has the type 'int'. It is convertible
> to any pointer type and to any pointer-to-member type, but the
> compiler needs to know what the destination type is. It cannot
> know from the type from the expression you supplied, that's why it
> complains.
> ...

Strictly speaking, the compiler cannot deduce the type simply because
default arguments do not participate in template argument deduction at
all. The fact that integral constant '0' is convertible to many
different pointer types plays no role here.

--
Best regards,
Andrey Tarasevich

Re: Default function parameter values
 

Tim Frink wrote:
>> Because the compiler needs to know what 'T' is. It cannot deduce
>> the type from an integral expression ('0'). You need to tell it
>> what the type T is by supplying the template argument, like
>>
>> objectA.foo<sometype>(string("test"));
>
> But why can I invoke the function with:
>
> objectA.foo( string( "test" ), objectA, &A::funcPtr );

You cannot. The second argument must have a pointer type. Your 'objectA'
is not a pointer (and user-defined conversion operators, if any, will
not be considered in this case). Therefore, the above invocation is
ill-formed.

You can invoke it as follows

objectA.foo( string( "test" ), &objectA, &A::funcPtr );

(note the extra '&').

> where I do not specify the type of T either?
> Where does the compiler get to information what type T is?

It is called "function template argument deduction". When you specify an
explicit argument for the corresponding parameter, the compiler will
analyze the type of the argument you supplied, and _deduce_ the meaning
of 'T' from that type. For example, in the above (corrected) call, the
meaning of 'T' can be deduced from either the second or the third
argument. It has to be 'A'.

When you don't supply explicit arguments, the complier has nothing to
deduce the meaning of 'T' from. (Default arguments are ignored by the
deduction process).

--
Best regards,
Andrey Tarasevich

Re: Default function parameter values
 

> You can invoke it as follows
>
> objectA.foo( string( "test" ), &objectA, &A::funcPtr );
>
> (note the extra '&').

You are completely right, I just missed the '&' in my post.


>
> When you don't supply explicit arguments, the complier has nothing to
> deduce the meaning of 'T' from. (Default arguments are ignored by the
> deduction process).

Thank you, now it's clear.