template function

[saneman]

I have this template function:

template <typename V, typename I>
void increasing_sequence(I p, I q) {
int j = 1;
while (p != q) {
*p = V(j);
j++; // increasing
++p;
}
}

I then call it with:

increasing_sequence<int>(v.begin(), v.end());

where v could be a container like std::vector. But why is that call
legal? I only specify one template parameter and the function has two.

 

[Victor Bazarov]

I have this template function:

template <typename V, typename I>
void increasing_sequence(I p, I q) {
int j = 1;
while (p != q) {
*p = V(j);
j++; // increasing
++p;
}
}

I then call it with:

increasing_sequence<int>(v.begin(), v.end());

where v could be a container like std::vector. But why is that call
legal? I only specify one template parameter and the function has two.

The other template argument ('I') is _deduced_ from the function call.
Since you pass both the first and second arguments of the same type,
the compiler can deduce it (and it is 'std::vector<???>::iterator',
unless your vector is 'const', in which case it's the const_iterator).

V

 

[Ian Collins]

The other template argument ('I') is _deduced_ from the function call.
Since you pass both the first and second arguments of the same type,
the compiler can deduce it (and it is 'std::vector<???>::iterator',
unless your vector is 'const', in which case it's the const_iterator).

Also note the first parameter can not be deduced.

 

[red floyd]

Also note the first parameter can not be deduced.

Well, not directly, but there's always
std::iterator_traits<I>::value_type if you don't want to have to specify
the first parameter.

 

[Kai-Uwe Bux]

Well, not directly, but there's always
std::iterator_traits<I>::value_type if you don't want to have to specify
the first parameter.

Except, in this case, the type I can very well be an output iterator. Then,
std::iterator_traits<I>::value_type could be void and not really convey the
needed information.


Best

Kai-Uwe Bux

 

[Ian Collins]

Well, not directly, but there's always
std::iterator_traits<I>::value_type if you don't want to have to specify
the first parameter.

Assuming that's what the OP wants...