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Question on type deduction

 
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francis_r
Guest
PostPosted: Wed Jul 12, 2006 1:42 am    Post subject: Question on type deduction Reply with quote


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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Carl Barron
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PostPosted: Thu Jul 13, 2006 3:29 am    Post subject: Re: Question on type deduction Reply with quote


In article <1152610779.648325.84140 (AT) 75g2000cwc (DOT) googlegroups.com>,
francis_r <francis.rammeloo (AT) gmail (DOT) com> wrote:

Quote:
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 );
}

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?

template <class R,class T

R exec_function(R (*f)(T &),T &x)
{
return f(x);
}

template <class R,class T>
R exec_function(R (*f)(T),T x)
{
return f(x);
}

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kernel0
Guest
PostPosted: Thu Jul 13, 2006 3:30 am    Post subject: Re: Question on type deduction Reply with quote


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:
Quote:
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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Owen
Guest
PostPosted: Thu Jul 13, 2006 3:32 am    Post subject: Re: Question on type deduction Reply with quote
francis_r:
Quote:

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


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


[...]


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Ian Wakeling
Guest
PostPosted: Thu Jul 13, 2006 3:50 am    Post subject: Re: Question on type deduction Reply with quote
francis_r wrote:

Quote:
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


Look at the types involved in the call that doesn't work:
Quote:
ExecuteFunction( Augment, theNumber ); // <-- compiler error
Augment is a pointer to a function that takes an int& and returns void.

theNumber is an int.

So, the generated instance of ExecuteFunction will look like this:

void ExecuteFunction( void (*inFunction)( int& ), int inArg1 )
{
return inFunction( inArg1 );
}

Hopefully the problem is now obvious: inArg1 is an int, which cannot be
converted to an int& for the call to inFunction.

To be honest, the use of an int& is not really good style anyway: given the
calling site, how is the poor maintenance programmer supposed to know that
calling ExecuteFunction() is going to alter the value of theNumber? Yes,
they can work it out by looking at the definition of Augment, but you are
just setting traps for the unwary, which won't earn you any friends when
the pressure is on to iron out the last few bugs and ship the product.

The good news is that if you remove the use of int&, you not only make the
code more obvious, you solve the compilation problem as well. Either
replace the int& with an int*, so that the calling site becomes

ExecuteFunction( Augment, &theNumber );

or adopt a more value based approach (which will sit better with the STL)
and change Augment to return the altered value instead:

int Augment( int num )
{
return( num + 1 );
}

This also yields a more readable calling site:
int theNumber = 10;
theNumber = ExecuteFunction( Augment, theNumber );

Ian


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Przemyslaw Szymanski
Guest
PostPosted: Thu Jul 13, 2006 4:03 am    Post subject: Re: Question on type deduction Reply with quote
francis_r wrote:
Quote:
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?

The problem is that argument deduction mechanism is trying to deduce
the actual type of Type1 template parameter from both actual parameters
of ExecuteFunction instantiation. From the first argument it is 'int &'
but from the second it's 'int', so the deduction fails. You could try
to introduce a nondeducible context for the second parameter which will
limit deduction to the first argument only:

template <typename T>
struct wrapper
{
typedef T type;
};

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

Regards,
Przemek.


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