Re: Question on type deduction

From:

"kernel0" <thewhiteapple@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

12 Jul 2006 18:30:46 -0400

Message-ID:

<1152688971.701507.307420@75g2000cwc.googlegroups.com>

change template function signature

from
------------------------------------------------------------------------------------------------------------
template< typename ReturnType, typename Type1 >
ReturnType ExecuteFunction( ReturnType (*inFunction)( Type1 ), Type1
inArg1 )

to

template< typename ReturnType, typename Type1 >
ReturnType ExecuteFunction( ReturnType (*inFunction)( Type1& ), Type1
inArg1 )

francis_r wrote:

Following very simplified code will illustrate my problem:

void Augment( int& outNumber )
{
    outNumber++;
}

template< typename ReturnType, typename Type1 >
ReturnType ExecuteFunction( ReturnType (*inFunction)( Type1 ), Type1
inArg1 )
{
    return inFunction( inArg1 );
}

int main()
{
    int theNumber = 10;
    ExecuteFunction( Augment, theNumber ); // <-- compiler error
    ExecuteFunction< void, int& >( Augment, theNumber ); // compiles OK
}

Compiler error goes as follows:
function call
ExecuteFunction({lval} void (int &), {lval} int)' does not match
'ExecuteFunction<...>(__T0 (*)(__T1), __T1)'
on line 435 ExecuteFunction( Augment, theNumber );

It seems that the second argument (theNumber) is not passed by
reference even though the Augment requests int& explicitely.

Is there a way I can rewrite the function ExecuteFunction so that I can
call it without having to explicitely specify it's template parameters?
If so, how?

Thanks in advance for all helpful feedback.

Kind regards,
Francis

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