c++ - c++ template function

Question

I have following code:

void myfunc() 
{ 
} 
template <typename T> 
void check() 
{ 
} 

template <typename T> 
void checkT (T) 
{ 
    check<T>(); 
} 

and so if I have in main function a call to checkT(myfunc) then that compiles, but if I have check<myfunc>() that doesn't work although it directly calls the first version. Can you please explain why it is so? The error is

error: no matching function for call to 'check()'

Thanks!

Answer 1

This is because myfunc is a value expression, not a type. You could do

check<decltype(myfunc)>();

though, or equivalently:

check<void(void)>();

See it live on http://liveworkspace.org/code/2ANEre$0

---

PS. In reply to the comment, I sense a bit of confusion between the function formal parameter and a template type parameter. Make it more explicit by writing:

template <typename T> 
void checkT (T somevalue) 
{ 
    check<T>();  // note: somevalue is never used!
} 

Answer 2

In the first instance checkT(myfunc) it is able to deduce the type, checkT is really identical to checkT( T value ) and so you are passing in value and T is being deduced. In the second case you are not supplying a type you can alter it like so to work:

check<decltype(myfunc)>() ;

you are actually supplying a value where you need a type in this case void(void).

Answer 3

checkT(myfunc) works because myfunc is a value. But the second fails because it is not a type, which the template parameters require. For example,

void checkT(T) 

is the same as

void checkT(T t)

so that means the function passed is an object of type T. That is, t is the object and T is the type. In the template parameters of check, it requires an explicit specification of the type, not the object. So thus passing in an object would raise a compilation error. Just like passing in the number 5 where the explicit type int is expected.

You can use it as a type by wrapping it in a decltype expression:

check<decltype(myfunc)>();
//    ^^^^^^^^^^^^^^^^