How to pass a template struct as parameter C++

Question

I got this template:

template <typename T>
struct List {
    struct node {
        T key;
        node* prev;
        node* next;
    };
    node* first = nullptr;
    node* last = nullptr;
    // functions
};

I need to create a function OUTSIDE of the struct that takes a List as a parameter. I tried something but doesnt work

void palindrome(List<typename T> L)

Proper way would be implement for list: `begin`, `end`, `rbegin`, `rend` and so on, then use something like this: https://stackoverflow.com/a/71437826/1387438 but this might be to hard for you for now. — Marek R, Mar 11 '22 at 14:11
Not only classes but also functions can be templates, basically exactly for your scenario: `template void palindrome(List t) {}`. — Peter - Reinstate Monica, Mar 11 '22 at 14:13
Alternatively, if the function caters just to one specific list element type, a specific List instantiation is basically an ordinary type, e.g. `void f(List cl) {}`. (That function is not a template. It does not depend on any type that can vary at compile time.) — Peter - Reinstate Monica, Mar 11 '22 at 14:21

score 2 · Answer 1 · answered Mar 11 '22 at 14:18

Your List is a class template (not template class / struct, it is not a class). You cannot pass a template to a function. You can only pass objects of some type to a function. List is not a type. List<double> for example is a type. So you could do for example:

void palindrome(List<double> L);

However, I suppose you want to pass any instantation of List to the function, eg a List<double> or a List<int> etc. Then you can make the function itself a function template:

template <typename T>
template void palindrome(List<T> l);

Note that this is not a function that accepts any instanatiation of List. It is just a template, and calling it with a List<int> or a List<double> will instantiate two distinct functions.

Sometimes it is simpler to be more generic, and you can do as well

template <typename L>
template void palindrome(L l);

You should also fix problem with list copy. – Marek R Mar 11 '22 at 15:04 — Marek R, Mar 11 '22 at 15:04

score 1 · Answer 2 · answered Mar 11 '22 at 14:17

In your function the typename T is unbound. The compiler has no idea what to do with that. You need to create a template function like this:

template<typename T> void palindrome(List<T> t) {}

The compiler will then deduce the typename T when you use the function.

Jason · Answer 3 · 2022-03-11T15:12:10.803

List is a class template which is diferent from a class type. A class type would be List<int> or a List<double> etc.

You can make palindrome a function template that has a parameter of type List<T> for some type T, as shown below.

//a function template that has parameter of type List<T> 
template<typename T>
void palindrome(List<T> L)
{
    std::cout<<"palindrome called"<<std::endl;
}
int main()
{
    List<int> l1;
    //call 
    palindrome(l1);//T will be deduced to int 
    
    List<double> l2;
    //call 
    palindrome(l2); //T will be deduced to double
    return 0;
}

Demo

How to pass a template struct as parameter C++

3 Answers3