Passing multiple functions to a template
0
The following code doesn't compile:
template< typename Fn1, typename Fn2 >
bool templateFunctionOne( Fn1&& fn1, Fn2&& fn2 )
{
int v = 5;
fn1( v );
return fn2( v );
}
template < typename Fn1, typename Fn2 >
bool templateFunctionTwo( Fn1&& fn1, Fn2&& fn2 )
{
std::future< bool > tk( std::async( std::launch::async,
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2) ) );
return tk.get();
}
bool printThis( int value )
{
cout << "this value = "
<< value
<< endl;
return true;
}
bool printThat( int value )
{
cout << "that value = "
<< value
<< endl;
return true;
}
int main()
{
auto func1 = std::bind( &printThis, std::placeholders::_1 );
auto func2 = std::bind( &printThat, std::placeholders::_2 );
return templateFunctionTwo( func1, func2 );
}
I'm getting errors like these:
error: no match for call to '(std::_Bind<boo (*(std::_Placeholder<2>))(int)>) (int&)'
return fn2( thatValue );
~~~^~~~~~~~~~~~~
template argument deduction/substitution failed:
candidate: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}; _Result = _Result; _Functor = bool (*)(int); _Bound_args = {std::_Placeholder<2>}]
operator()(_Args&&... --args) cont
^~~~~~~~
I know it has something to do with passing multiple function pointers to the template, but just don't know where and what. Could anyone help point out where my mistakes are please?
c++ templates
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
add a comment |
0
The following code doesn't compile:
template< typename Fn1, typename Fn2 >
bool templateFunctionOne( Fn1&& fn1, Fn2&& fn2 )
{
int v = 5;
fn1( v );
return fn2( v );
}
template < typename Fn1, typename Fn2 >
bool templateFunctionTwo( Fn1&& fn1, Fn2&& fn2 )
{
std::future< bool > tk( std::async( std::launch::async,
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2) ) );
return tk.get();
}
bool printThis( int value )
{
cout << "this value = "
<< value
<< endl;
return true;
}
bool printThat( int value )
{
cout << "that value = "
<< value
<< endl;
return true;
}
int main()
{
auto func1 = std::bind( &printThis, std::placeholders::_1 );
auto func2 = std::bind( &printThat, std::placeholders::_2 );
return templateFunctionTwo( func1, func2 );
}
I'm getting errors like these:
error: no match for call to '(std::_Bind<boo (*(std::_Placeholder<2>))(int)>) (int&)'
return fn2( thatValue );
~~~^~~~~~~~~~~~~
template argument deduction/substitution failed:
candidate: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}; _Result = _Result; _Functor = bool (*)(int); _Bound_args = {std::_Placeholder<2>}]
operator()(_Args&&... --args) cont
^~~~~~~~
I know it has something to do with passing multiple function pointers to the template, but just don't know where and what. Could anyone help point out where my mistakes are please?
c++ templates
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
3
Your secondbindcall creates a function object expecting two arguments; it ignores the first and passes the second toprintThat. ButtemplateFunctionOnecalls it with one argument. You likely meantstd::placeholders::_1in both places. In fact, you can dropbindcalls and just passprintThisandprintThatdirectly.
– Igor Tandetnik
Nov 25 '18 at 17:49
auto func2 = std::bind( &printThat, std::placeholders::_2 );->auto func2 = std::bind( &printThat, std::placeholders::_1 );
– Matthieu Brucher
Nov 25 '18 at 17:50
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05
add a comment |
0
0
0
The following code doesn't compile:
template< typename Fn1, typename Fn2 >
bool templateFunctionOne( Fn1&& fn1, Fn2&& fn2 )
{
int v = 5;
fn1( v );
return fn2( v );
}
template < typename Fn1, typename Fn2 >
bool templateFunctionTwo( Fn1&& fn1, Fn2&& fn2 )
{
std::future< bool > tk( std::async( std::launch::async,
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2) ) );
return tk.get();
}
bool printThis( int value )
{
cout << "this value = "
<< value
<< endl;
return true;
}
bool printThat( int value )
{
cout << "that value = "
<< value
<< endl;
return true;
}
int main()
{
auto func1 = std::bind( &printThis, std::placeholders::_1 );
auto func2 = std::bind( &printThat, std::placeholders::_2 );
return templateFunctionTwo( func1, func2 );
}
I'm getting errors like these:
error: no match for call to '(std::_Bind<boo (*(std::_Placeholder<2>))(int)>) (int&)'
return fn2( thatValue );
~~~^~~~~~~~~~~~~
template argument deduction/substitution failed:
candidate: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}; _Result = _Result; _Functor = bool (*)(int); _Bound_args = {std::_Placeholder<2>}]
operator()(_Args&&... --args) cont
^~~~~~~~
I know it has something to do with passing multiple function pointers to the template, but just don't know where and what. Could anyone help point out where my mistakes are please?
c++ templates
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
The following code doesn't compile:
template< typename Fn1, typename Fn2 >
bool templateFunctionOne( Fn1&& fn1, Fn2&& fn2 )
{
int v = 5;
fn1( v );
return fn2( v );
}
template < typename Fn1, typename Fn2 >
bool templateFunctionTwo( Fn1&& fn1, Fn2&& fn2 )
{
std::future< bool > tk( std::async( std::launch::async,
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2) ) );
return tk.get();
}
bool printThis( int value )
{
cout << "this value = "
<< value
<< endl;
return true;
}
bool printThat( int value )
{
cout << "that value = "
<< value
<< endl;
return true;
}
int main()
{
auto func1 = std::bind( &printThis, std::placeholders::_1 );
auto func2 = std::bind( &printThat, std::placeholders::_2 );
return templateFunctionTwo( func1, func2 );
}
I'm getting errors like these:
error: no match for call to '(std::_Bind<boo (*(std::_Placeholder<2>))(int)>) (int&)'
return fn2( thatValue );
~~~^~~~~~~~~~~~~
template argument deduction/substitution failed:
candidate: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}; _Result = _Result; _Functor = bool (*)(int); _Bound_args = {std::_Placeholder<2>}]
operator()(_Args&&... --args) cont
^~~~~~~~
I know it has something to do with passing multiple function pointers to the template, but just don't know where and what. Could anyone help point out where my mistakes are please?
c++ templates
c++ templates
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
edited Nov 25 '18 at 17:48
Matthieu Brucher
16.6k32243
16.6k32243
asked Nov 25 '18 at 17:45
markmark
325
asked Nov 25 '18 at 17:45
markmark
325
asked Nov 25 '18 at 17:45
markmark
325
325
3
Your secondbindcall creates a function object expecting two arguments; it ignores the first and passes the second toprintThat. ButtemplateFunctionOnecalls it with one argument. You likely meantstd::placeholders::_1in both places. In fact, you can dropbindcalls and just passprintThisandprintThatdirectly.
– Igor Tandetnik
Nov 25 '18 at 17:49
auto func2 = std::bind( &printThat, std::placeholders::_2 );->auto func2 = std::bind( &printThat, std::placeholders::_1 );
– Matthieu Brucher
Nov 25 '18 at 17:50
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05
add a comment |
3
Your secondbindcall creates a function object expecting two arguments; it ignores the first and passes the second toprintThat. ButtemplateFunctionOnecalls it with one argument. You likely meantstd::placeholders::_1in both places. In fact, you can dropbindcalls and just passprintThisandprintThatdirectly.
– Igor Tandetnik
Nov 25 '18 at 17:49
auto func2 = std::bind( &printThat, std::placeholders::_2 );->auto func2 = std::bind( &printThat, std::placeholders::_1 );
– Matthieu Brucher
Nov 25 '18 at 17:50
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05
3
3
Your second
bind call creates a function object expecting two arguments; it ignores the first and passes the second to printThat. But templateFunctionOne calls it with one argument. You likely meant std::placeholders::_1 in both places. In fact, you can drop bind calls and just pass printThis and printThat directly.– Igor Tandetnik
Nov 25 '18 at 17:49
Your second
bind call creates a function object expecting two arguments; it ignores the first and passes the second to printThat. But templateFunctionOne calls it with one argument. You likely meant std::placeholders::_1 in both places. In fact, you can drop bind calls and just pass printThis and printThat directly.– Igor Tandetnik
Nov 25 '18 at 17:49
auto func2 = std::bind( &printThat, std::placeholders::_2 ); -> auto func2 = std::bind( &printThat, std::placeholders::_1 );– Matthieu Brucher
Nov 25 '18 at 17:50
auto func2 = std::bind( &printThat, std::placeholders::_2 ); -> auto func2 = std::bind( &printThat, std::placeholders::_1 );– Matthieu Brucher
Nov 25 '18 at 17:50
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05
add a comment |
1 Answer
1
active
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votes
0
Stop using std bind.
return templateFunctionTwo( &printThis, &printThat );
every use of it in the above code makes your code less clear and more error prone.
This is also a mess:
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2)
why not
[fn1=std::forward<Fn1>(fn1),fn2=std::forward<Fn2>(fn2)]()mutable{ templateFunctionOne(std::move(fn1), std::move(fn2)); }
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
add a comment |
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1 Answer
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0
Stop using std bind.
return templateFunctionTwo( &printThis, &printThat );
every use of it in the above code makes your code less clear and more error prone.
This is also a mess:
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2)
why not
[fn1=std::forward<Fn1>(fn1),fn2=std::forward<Fn2>(fn2)]()mutable{ templateFunctionOne(std::move(fn1), std::move(fn2)); }
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
add a comment |
0
Stop using std bind.
return templateFunctionTwo( &printThis, &printThat );
every use of it in the above code makes your code less clear and more error prone.
This is also a mess:
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2)
why not
[fn1=std::forward<Fn1>(fn1),fn2=std::forward<Fn2>(fn2)]()mutable{ templateFunctionOne(std::move(fn1), std::move(fn2)); }
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
add a comment |
0
0
0
Stop using std bind.
return templateFunctionTwo( &printThis, &printThat );
every use of it in the above code makes your code less clear and more error prone.
This is also a mess:
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2)
why not
[fn1=std::forward<Fn1>(fn1),fn2=std::forward<Fn2>(fn2)]()mutable{ templateFunctionOne(std::move(fn1), std::move(fn2)); }
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
Stop using std bind.
return templateFunctionTwo( &printThis, &printThat );
every use of it in the above code makes your code less clear and more error prone.
This is also a mess:
&templateFunctionOne< typename std::remove_reference<Fn1>::type,
typename std::remove_reference<Fn2>::type >,
std::forward<Fn1>(fn1),
std::forward<Fn2>(fn2)
why not
[fn1=std::forward<Fn1>(fn1),fn2=std::forward<Fn2>(fn2)]()mutable{ templateFunctionOne(std::move(fn1), std::move(fn2)); }
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
answered Nov 25 '18 at 18:29
Yakk - Adam NevraumontYakk - Adam Nevraumont
187k21199384
187k21199384
add a comment |
add a comment |
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3
Your second
bindcall creates a function object expecting two arguments; it ignores the first and passes the second toprintThat. ButtemplateFunctionOnecalls it with one argument. You likely meantstd::placeholders::_1in both places. In fact, you can dropbindcalls and just passprintThisandprintThatdirectly.– Igor Tandetnik
Nov 25 '18 at 17:49
auto func2 = std::bind( &printThat, std::placeholders::_2 );->auto func2 = std::bind( &printThat, std::placeholders::_1 );– Matthieu Brucher
Nov 25 '18 at 17:50
Thank you Mathieu, that was it!! Greatly appreciate it!
– mark
Nov 25 '18 at 17:52
All the merit rests with @IgorTandetnik
– Matthieu Brucher
Nov 25 '18 at 17:55
Works best with return 0;
– QuentinUK
Nov 25 '18 at 18:05