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

Sure.  'objectA' has a particular type.  '&A::funcPtr' has it too.

V
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