[boost::function] problem with void type as argument type

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Re: [Boost-users] [bind] static...

Seweryn Habdank-Wojewódzki

26 Mar 2007 26 Mar '07

6:31 p.m.

Hi! I have small piece of code and I do not know why it is not compiling - proper instantiation of function is not created. It means, why boost::function can not deduct types (code below). The problem is when type U is void. Can any one help me with it? Regards. template < typename T, typename U > void test ( boost::function & f ) { f(); } int f3() { std::cout << "f3(void)" << std::endl; return 3; } void f4() { std::cout << "f4(void)" << std::endl; return;// 4; } int main() { boost::function h = &f3; boost::function i = &f4; test (h); test (i); } -- |\/\/| Seweryn Habdank-Wojewódzki \/\/

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Douglas Gregor

26 Mar 26 Mar

6:46 p.m.

New subject: [boost::function] problem with void type as argument type

On Mon, 2007-03-26 at 20:31 +0200, Seweryn Habdank-Wojewódzki wrote:

...

Hi!

I have small piece of code and I do not know why it is not compiling - proper instantiation of function is not created.

It means, why boost::function can not deduct types (code below).

The problem is when type U is void. Can any one help me with it?

Regards.

template < typename T, typename U > void test ( boost::function & f ); [snip] boost::function h = &f3; boost::function i = &f4;

The problem is that "int(void)" is just an alternative spelling for "int()", i.e., a function type for a function that takes no parameters but returns an "int". A function type int() cannot match a template parameter T(U), because there is nothing to match the "U". Cheers, Doug

Seweryn Habdank-Wojewódzki

7:39 p.m.

New subject: [boost::function] problem with void type as argument type

Hi! Douglas Gregor wrote:

...

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template < typename T, typename U > void test ( boost::function & f ); [snip] boost::function h = &f3; boost::function i = &f4;

The problem is that "int(void)" is just an alternative spelling for "int()", i.e., a function type for a function that takes no parameters but returns an "int". A function type int() cannot match a template parameter T(U), because there is nothing to match the "U".

Thanks for reply. However still I do not know how to proceed in that case, because functions can not be partially specialized. Is there any opportunity to prepare more less such a function which will automatically detects arguments of boost::function template? Best regards. -- |\/\/| Seweryn Habdank-Wojewódzki \/\/

Douglas Gregor

7:46 p.m.

New subject: [boost::function] problem with void type as argument type

On Mon, 2007-03-26 at 21:39 +0200, Seweryn Habdank-Wojewódzki wrote:

...

Hi!

Douglas Gregor wrote:

...
...
template < typename T, typename U > void test ( boost::function & f ); [snip] boost::function h = &f3; boost::function i = &f4;

The problem is that "int(void)" is just an alternative spelling for "int()", i.e., a function type for a function that takes no parameters but returns an "int". A function type int() cannot match a template parameter T(U), because there is nothing to match the "U".

Thanks for reply. However still I do not know how to proceed in that case, because functions can not be partially specialized. Is there any opportunity to prepare more less such a function which will automatically detects arguments of boost::function template?

You can write multiple overloads of your "test" function, each of which accepts boost::function instances with a different number of function arguments: template<typename R> void test (boost::function& f); template void test (boost::function& f); template void test (boost::function& f); template void test (boost::function& f); Tedious, unfortunately, but it works. Cheers, Doug

Seweryn Habdank-Wojewódzki

8:28 p.m.

New subject: [boost::function] problem with void type as argument type

Hi! Douglas Gregor wrote:

...

...
Thanks for reply. However still I do not know how to proceed in that case, because functions can not be partially specialized. Is there any opportunity to prepare more less such a function which will automatically detects arguments of boost::function template?

You can write multiple overloads of your "test" function, each of which accepts boost::function instances with a different number of function arguments:

template<typename R> void test (boost::function& f);

template void test (boost::function& f);

template void test (boost::function& f);

template void test (boost::function& f);

Tedious, unfortunately, but it works.

Ok. It is long, but good enough. Thanks and kind regards. -- |\/\/| Seweryn Habdank-Wojewódzki \/\/

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Douglas Gregor
Seweryn Habdank-Wojewódzki