Re: Question on type deduction

From:

Thomas Maeder <maeder@glue.ch>

Newsgroups:

comp.lang.c++.moderated

Date:

12 Jul 2006 18:28:41 -0400

Message-ID:

<m2lkqzz742.fsf@glue.ch>

"francis_r" <francis.rammeloo@gmail.com> writes:

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.

The two decisions take place in the reverse order. First, the types of

the arguments are established, then the available ExecuteFunction

overloads are attempted to be matched.

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?

You could add an overload

template <typename ReturnType, typename Type1>

ReturnType ExecuteFunction(ReturnType (*inFunction)(Type1 &),

                           Type1 &inArg1)

{

  return inFunction(inArg1);

}

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