function templates doesn't support default template parameters?

[William Xu]

Compiling:

template <class T = int>
T foo(const T& t)
{}

int main(int argc, char *argv[])
{}

gcc complains:

,----
| /Users/william/repo/helloworlds/foo.cpp:2: error: default template
| arguments may not be used in function templates
`----

But I find in "TC++PL(3rd, special edition)" P.340, Bjarne is giving
function templates with default template parameters as examples.

If current standard doesn't support it, what is the reason here?

--
William

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2008-06-01 16:29, William Xu wrote:

Compiling:

template <class T = int>
T foo(const T& t)
{}

int main(int argc, char *argv[])
{}

gcc complains:

,----
| /Users/william/repo/helloworlds/foo.cpp:2: error: default template
| arguments may not be used in function templates
`----

But I find in "TC++PL(3rd, special edition)" P.340, Bjarne is giving
function templates with default template parameters as examples.

If current standard doesn't support it, what is the reason here?

It is not supported in the current standard (14.1 §9), I do not know why.

--
Erik Wikström

[Kai-Uwe Bux]

Erik Wikström wrote:

On 2008-06-01 16:29, William Xu wrote:

>Compiling:

template <class T = int>
T foo(const T& t)
{}

int main(int argc, char *argv[])
{}

gcc complains:

,----
| /Users/william/repo/helloworlds/foo.cpp:2: error: default template
| arguments may not be used in function templates
`----

But I find in "TC++PL(3rd, special edition)" P.340, Bjarne is giving
function templates with default template parameters as examples.

If current standard doesn't support it, what is the reason here?

It is not supported in the current standard (14.1 §9), I do not know why.

I wonder what difference one should expect from

template < typename T = int >
T foo ( T const & arg );

and

template < typename T >
T foo ( T const & arg );

How would you use the default type? The type T would be deduced from the
argument anyway, wouldn't it?
Best

Kai-Uwe Bux

[William Xu]

Kai-Uwe Bux <jk********@gmx.netwrites:

How would you use the default type? The type T would be deduced from the
argument anyway, wouldn't it?

Consider the following example then. What if I want the default
comparision method to be Case_insensitive?

#include <iostream>

struct Case_insensitive
{
static bool eq(char c1, char c2) { return tolower(c1) == tolower(c2); }
};

struct Case_sensitive
{
static bool eq(char c1, char c2) { return c1 == c2; }
};

template <class Cmp>
bool eq(char c1, char c2)
{
return Cmp::eq(c1, c2);
}

int main(int argc, char *argv[])
{
char c1 = 'h', c2 = 'H';

/* These are okay. */
eq<Case_insensitive>(c1, c2) ;
eq<Case_sensitive>(c1, c2);

/* But how about this one ? */
/* eq(c1, c2); */

}

--
William

http://williamxu.net9.org

You know what they say -- the sweetest word in the English language is revenge.
-- Peter Beard

[Kai-Uwe Bux]

William Xu wrote:

Kai-Uwe Bux <jk********@gmx.netwrites:

>How would you use the default type? The type T would be deduced from the
argument anyway, wouldn't it?

Consider the following example then. What if I want the default
comparision method to be Case_insensitive?

#include <iostream>

struct Case_insensitive
{
static bool eq(char c1, char c2) { return tolower(c1) == tolower(c2); }
};

struct Case_sensitive
{
static bool eq(char c1, char c2) { return c1 == c2; }
};

template <class Cmp>
bool eq(char c1, char c2)
{
return Cmp::eq(c1, c2);
}

int main(int argc, char *argv[])
{
char c1 = 'h', c2 = 'H';

/* These are okay. */
eq<Case_insensitive>(c1, c2) ;
eq<Case_sensitive>(c1, c2);

/* But how about this one ? */
/* eq(c1, c2); */

}

Now, I can see your point.

On the other hand, it looks as though you want a default argument not a
default type:

#include <cctype>

bool case_insensitive ( char c1, char c2 ) {
return ( std::tolower(c1) == std::tolower(c2) );
}

bool case_sensitive ( char c1, char c2 ) {
return ( c1 == c2 );
}

typedef bool(* char_compare )(char,char);

bool eq ( char c1, char c2, char_compare comp = &case_sensitive ) {
return ( comp( c1, c2 ) );
}

int main ( void ) {
eq( 'c', 'h' );
}
Then again, I am not at all sure whether it is a good idea to have a default
in this case. I generally do not like magic hiding somewhere. Sooner or
later it is going to bite you. From that point of view, I prefer the
verbose version

eq< case_sensitive >( c1, c2 );

to

eq( c1, c2 );

Best

Kai-Uwe Bux

[James Kanze]

On Jun 1, 4:46 pm, Erik Wikström <[e-mail]:

On 2008-06-01 16:29, William Xu wrote:

Compiling:

template <class T = int>
T foo(const T& t)
{}

int main(int argc, char *argv[])
{}

gcc complains:

But I find in "TC++PL(3rd, special edition)" P.340, Bjarne
is giving function templates with default template
parameters as examples.

That surprises me a bit (but someone walked off with my copy of
the 3rd edition, so I can't verify it).

If current standard doesn't support it, what is the reason
here?

It is not supported in the current standard (14.1 §9), I do
not know why.

Probably because originally, function template arguments could
only be deduced, not explicitly specified, and a defauld
argument doesn't make sense in that case. For that matter,
given the original poster's example, when would the default
argument be used?

--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[Eric Pruneau]

"James Kanze" <ja*********@gmail.coma écrit dans le message de news:
ff**********************************...oglegroups.com...
On Jun 1, 4:46 pm, Erik Wikström <[e-mail]:

On 2008-06-01 16:29, William Xu wrote:

Compiling:

template <class T = int>
T foo(const T& t)
{}

int main(int argc, char *argv[])
{}

gcc complains:

But I find in "TC++PL(3rd, special edition)" P.340, Bjarne
is giving function templates with default template
parameters as examples.

That surprises me a bit (but someone walked off with my copy of
the 3rd edition, so I can't verify it).

If current standard doesn't support it, what is the reason
here?

It is not supported in the current standard (14.1 §9), I do
not know why.

Probably because originally, function template arguments could
only be deduced, not explicitly specified, and a defauld
argument doesn't make sense in that case. For that matter,
given the original poster's example, when would the default
argument be used?

--

From C++ Templates (Vandevoorde, Josuttis)

"When templates were originally added to the C++ language, explicit function
template arguments were not a valid construct. [...] Since then, however, it
is possible to specify explicitle function template arguments that cannot be
deduced. "

So the following should compile...

template <typename T1, typename T2 = int>
T2 count (T1 const& x);

Cause T2 cannot be deduced since it is the return parameter. However I tried
with intel c++ 9.1 and VS 2003 compilers and they give me an error...
--------
Eric Pruneau