Is there a 'functiondef' (similar to a 'typedef')?

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Peter Oliphant - 18 Oct 2005 00:08 GMT

I have defined a templated function:

template< typename dataT >
dataT Add_T( dataT x, dataT y ) { return (x+y) ; }

Now, I'd like to do something like:

typedef Add_T<int> Add_int ;

That is, turn the ugly " Add_T<int>( x, y )" into something like "Add_int(x,
y )" (where 'x' and 'y' are 'int's). This is how it's done with templated
class definitions. How is it done with templated functions?

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Carl Daniel [VC++ MVP] - 18 Oct 2005 03:13 GMT

> I have defined a templated function:
>
[quoted text clipped - 9 lines]
> done with templated class definitions. How is it done with templated
> functions?

You're almost out of luck - there's no such thing as a "funcdef".

What you could do though...

int (*Add_int)(int,int) = &Add_T<int>(int,int);

That declares a pointer to function taking (int,int) and returning int named
pAdd_T, and that pointer references the instantiation of your Add_T function
template.

You can call through a function pointer as-if it were a function:

int i = Add_int(3,5);

The downside is that using this technique will almost certainly prevent
Add_T<int> from being inlined when it's called through Add_int.

-cd

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Peter Oliphant - 18 Oct 2005 17:32 GMT

Thanx, Carl! : )

[==P==]

>> I have defined a templated function:
>>
[quoted text clipped - 28 lines]
>
> -cd

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Vladimir Nesterovsky - 20 Oct 2005 10:39 GMT

> template< typename dataT >
> dataT Add_T( dataT x, dataT y ) { return (x+y) ; }
[quoted text clipped - 6 lines]
> "Add_int(x, y )" (where 'x' and 'y' are 'int's). This is how it's done
> with templated class definitions. How is it done with templated functions?

I think you can call this function without angle brackets: Add_T(int_x,
int_y).

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Vladimir Nesterovsky

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Peter Oliphant - 20 Oct 2005 17:08 GMT

I tried it, and you're right! I guess that 'int' is what it assumes as a
default typename if not given one... Cool to know! : )

[==P==]

>> template< typename dataT >
>> dataT Add_T( dataT x, dataT y ) { return (x+y) ; }
[quoted text clipped - 10 lines]
> I think you can call this function without angle brackets: Add_T(int_x,
> int_y).

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Doug Harrison [MVP] - 20 Oct 2005 17:19 GMT

>I tried it, and you're right! I guess that 'int' is what it assumes as a
>default typename if not given one... Cool to know! : )

Not exactly. The template type parameters are deduced from the function
arguments. You only need to specify them yourself using the f<type>()
syntax if the template parameter isn't represented in the function argument
list, or it is, and it can't be deduced for some reason, such as calling
f(x, y), where both parameters have the type T, but x is (say) an int and y
is a long and thus aren't the same type. To fix that, you can say f<int>(x,
y).

Signature

Doug Harrison
Visual C++ MVP

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Peter Oliphant - 20 Oct 2005 21:11 GMT

Interesting, and thanks for the explanation! : )

But then, how do you explain that the following works:

int result = Add_T( 1, 2 ) ;

Here the parameters are not 'typed', since '1' and '2' could be 'long' or
'int' or 'unsigned' etc. Although I'm guessing that, as parameters,
constants of the form of an integer without further casting will be assumed
to be an 'int'. Or it could be grabbing the parameter type from the 'result'
type perhaps (since it is structured to be the same type in as out)?

[==P==]

>>I tried it, and you're right! I guess that 'int' is what it assumes as a
>>default typename if not given one... Cool to know! : )
[quoted text clipped - 9 lines]
> f<int>(x,
> y).

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Doug Harrison [MVP] - 20 Oct 2005 22:18 GMT

>Interesting, and thanks for the explanation! : )
>
[quoted text clipped - 7 lines]
>to be an 'int'. Or it could be grabbing the parameter type from the 'result'
>type perhaps (since it is structured to be the same type in as out)?

But 1 and 2 are integer literals that have the type int. For more on this,
see:

C++ Integer Constants
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vclang/html/_pl
uslang_C.2b2b_.Integer_Constants.asp?frame=true

There are even more obscure rules that aren't covered there, that determine
the type when the constant won't fit in its "preferred" type, and this
further depends on the form of the constant, e.g. decimal vs. hex, and
whether or not there's a suffix. Here's a page that covers that:

http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/index.jsp?topic=/com.ib
m.xlcpp8a.doc/language/ref/conref.htm

Signature

Doug Harrison
Visual C++ MVP

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Is there a 'functiondef' (similar to a 'typedef')?

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