Why can\'t I use a non-constant variable as a template argument in C ?

Non-Constant Template Argument Conundrum

When attempting to pass a non-constant variable as a template argument, such as i in the following code snippet, the compiler issues an error:

<code class="cpp">template <int a>
void modify() {}

for (int i = 0; i < 10; i++) {
    modify<i>(); // error: 'i' cannot appear in constant-expression
}</code>

Reason for the Error:

Templates are expanded during compilation, which requires their arguments to be evaluated at compile time. Since i is modified within the loop, the compiler cannot determine its value at compile time, leading to the error.

Alternative Implementation:

To achieve the desired iteration without altering the API interface, consider implementing the following:

<code class="cpp">#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>()();
}</code>

This version utilizes recursion to emulate iteration. By instantiating specialized template functions for each value (e.g., modify<0>, modify<1>, etc.), the code simulates the loop behavior from i=0 to i=9.

Non-Constant Template Argument Resolution:

To call modify with a variable argument VAR (determined by a functional computation), consider using a template function with variadic parameters, as follows:

<code class="cpp">template <typename T>
void modify(const T& x)
{ std::cout << "modify(" << x << ")" << std::endl; }

int main()
{
    auto VAR = ...; // computed from some functional process
    modify(VAR);
}</code>

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