template deduction of array size

[Gianni Mariani]

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

[Dave]

> template <unsigned N>

int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

I think not since neither Comeau nor VC++ 7.1 accept it (results below), but
I must admit I am perplexed as to why. I know that in general a nontype
parameter for a function template can be deduced, but there is clearly
something special about this case. I'll be interested to follow this
thread...

Note that if you change the parameter to bre reference to array, it works...
Maybe this can hint at why the case you presented fails...
Comeau:
"ComeauTest .c", line 11: error: no instance of function template "strn"
matches the
argument list
The argument types that you used are: (const char [5])
std::cout << strn("abcd");

VC++ 7.1:
d:\visual_c++\d ot_net\paramete r_test\paramete r_test\main.cpp (11) : error
C2784: 'int strn(const char [N])' : could not deduce template argument for
'const char [N]' from 'const char [5]'
d:\visual_c++\d ot_net\paramete r_test\paramete r_test\main.cpp (4) :
see declaration of 'strn'

[Dan W.]

On 03 Dec 2003 17:27:52 EST, Gianni Mariani <gi*******@mari ani.ws>
wrote:

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

Geez, I read an article in CUJ long time ago that dealt with this very
situation... I think you need to templetize with sizeof().

Hope I inspire you in the right general direction.

[Dave]

"Gianni Mariani" <gi*******@mari ani.ws> wrote in message
news:bq******** @dispatch.conce ntric.net...

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

One other thought: Both the parameter and the argument should decay to a
pointer. (The argument would not if it were reference, but that's not the
case here...). I wonder if this is somehow interfering with the deduction
process...

[Gianni Mariani]

Dave wrote:

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}
....

error: no matching function for call to `strn(const char[5])'

... One other thought: Both the parameter and the argument should decay to a
pointer. (The argument would not if it were reference, but that's not the
case here...). I wonder if this is somehow interfering with the deduction
process...

Yes, a reference seems to work, but why does the non reference version
fail ?

template <unsigned N>
int strn( const char ( & str )[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}
OK - I solved my problem.

[Rob Williscroft]

Gianni Mariani wrote in news:bq******** @dispatch.conce ntric.net:

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}
This is effectivly:

template < unsigned N >
int strn( char const *str )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

Yup the call needs to be:

std::cout << strn< 5 >( "abcd" );

Where 5 can be any compile time integral constant you want.

template < unsigned N >
unsigned strn( char const (&str)[ N ] )
{
return N;
}

Or get creative:

template < typename T, unsigned N >
unsigned countof( T const (&array)[ N ] )
{
return N;
}

HTH.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Dan W.]

On Wed, 03 Dec 2003 17:42:53 -0500, Dan W. <da**@raytron-controls.com>
wrote:

On 03 Dec 2003 17:27:52 EST, Gianni Mariani <gi*******@mari ani.ws>
wrote:

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'

Geez, I read an article in CUJ long time ago that dealt with this very
situation... I think you need to templetize with sizeof().

Hope I inspire you in the right general direction.

What's wrong with sizeof() itself?

[Dan W.]

On 03 Dec 2003 18:11:37 EST, Gianni Mariani <gi*******@mari ani.ws>
wrote:

Dave wrote:

Does anyone know if this is supposed to work ?

template <unsigned N>
int strn( const char str[N] )
{
return N;
}
...

error: no matching function for call to `strn(const char[5])'

..

One other thought: Both the parameter and the argument should decay to a
pointer. (The argument would not if it were reference, but that's not the
case here...). I wonder if this is somehow interfering with the deduction
process...

Yes, a reference seems to work, but why does the non reference version
fail ?

template <unsigned N>
int strn( const char ( & str )[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}
OK - I solved my problem.

I think that in the case of the array, the number in the brackets is
never passed to the function at runtime; --i.e.: it is equivalent to
passing a pointer, and the number in the brackets is just candy for
your eyes only; whereas a reference is a unique type.

[Dan W.]

>

I think that in the case of the array, the number in the brackets is
never passed to the function at runtime; --i.e.: it is equivalent to
passing a pointer, and the number in the brackets is just candy for
your eyes only; whereas a reference is a unique type.

Of course! int[5] is not a compile-time type, but a reference to
int[5] IS a type.

Just a C clingon.

[Dave]

> > template <unsigned N>

int strn( const char str[N] )
{
return N;
}

#include <iostream>

int main()
{
std::cout << strn( "abcd" );
}

error: no matching function for call to `strn(const char[5])'
I think not since neither Comeau nor VC++ 7.1 accept it (results below),

but I must admit I am perplexed as to why. I know that in general a nontype
parameter for a function template can be deduced, but there is clearly
something special about this case. I'll be interested to follow this
thread...

Note that if you change the parameter to bre reference to array, it works... Maybe this can hint at why the case you presented fails...
Comeau:
"ComeauTest .c", line 11: error: no instance of function template "strn"
matches the
argument list
The argument types that you used are: (const char [5])
std::cout << strn("abcd");

VC++ 7.1:
d:\visual_c++\d ot_net\paramete r_test\paramete r_test\main.cpp (11) : error
C2784: 'int strn(const char [N])' : could not deduce template argument for
'const char [N]' from 'const char [5]'
d:\visual_c++\d ot_net\paramete r_test\paramete r_test\main.cpp (4) :
see declaration of 'strn'

Has a conclusion been reached on *why* the original code (shown at the top)
fails to compile? Even though we have determined that using a reference
fixes the problem, it is still unclear to me why the original program fails
to compile...