Question

I am working with a library which exposes an interface to work with. One of the functions of this library is like this :

template <int a>
void modify(){}

I have to modify parameters from 1 to 10 i.e. call modify with with template arguments from 1 to 10. For that I wrote this code (a basic version of code, actual code is much larger).

for(int i = 0; i < 10; i++) {
    modify<i>(); // <<< error on this line
}

On compilation I receive the following error

error: 'i' cannot appear in constant-expression

After going through some links on the internet, I came to know that I cannot pass any value as template argument which is not evaluated at compile time.
My question are as follows:

1. Why can’t compiler evaluate i at compile time?
2. Is there any other to achieve the objective I am trying to achieve without changing the API interface?

---

There is another thing I want to do. Call modify as modify where VAR is the output of some functional computation. How can I do that?

Answer 1

What is the value of i (that is not a constant) at compile time? There is no way to answer unless executing the loop. But executing is not “compiling”
Since there is no answer, the compiler cannot do that.

Templates are not algorithm to be executed, but macros to be expanded to produce code.
What you can do is rely on specialization to implement iteration by recursion, like here:

#include <iostream>

template<int i>
void modify()
{ std::cout << "modify<"<<i<<">"<< std::endl; }

template<int x, int to>
struct static_for
{
    void operator()() 
    {  modify<x>();  static_for<x+1,to>()(); }
};

template<int to>
struct static_for<to,to>
{
    void operator()() 
    {}
};


int main()
{
    static_for<0,10>()();
}

Note that, by doing this, you are, in fact, instantiating 10 functions named
modify<0> … modify<9>, called respectively by static_for<0,10>::operator() … static_for<9,10>::operator().

The iteration ends because static_for<10,10> will be instantiated from the specialization that takes two identical values, that does nothing.