从函数模板错误返回修改后的值

Question

My code is something like this:

   // ... code

   template <int i>
   int modifyparameter()
   {
     i = i++;
     return i;
   }

   // ... some more code

   int main()
   {
        int answer = modifyparameter<5>();
        cout << answer; //expecting 6
   }

But I am getting errors. What have I done wrong?

Answer 1

i is the name of an int value , and you cannot modify values. You probably want one of these:

template <typename Number>
Number functional(Number x)
{
    return x + 1;
}

template <typename Number>
Number& side_effect(Number& x)
{
    return ++x;
}

Answer 2

i is not an lvalue , so i++ is illegal.

14.1/5 says.

A non-type non-reference template-parameter is not an lvalue. It shall not be assigned to or in any other way have its value changed.

Post-increment operator ( ++ ) requires an lvalue as an operand.

Answer 3

Try:

template <int i>
int modifyparameter()
{
  int copy_of_i = i;
  copy_of_i++;
  return copy_of_i;
}

There's nothing you can do to modify i itself, template parameters are compile-time constants. However, you could make it act as if I was being modified by using a variable with static lifetime:

template <int i>
int modifyparameter()
{
  static int copy_of_i = i;
  copy_of_i++;
  return copy_of_i;
}

Answer 4

While it's not asked for, this seems to cry out for a compile-time solution:

template< int I >
struct increment { static const int result = I+1; };

std::cout << increment<41>::result << '\n';

That struct is what's called a meta-function. Yes, it's a struct, but it's used (at compile-time) like a function: you call it, passing a parameter, and you get back a result. The syntax is hilarious, but that's because nobody actually planned to do this; the possibility was discovered (more or less) by accident.

Doing this at compile-time has the advantage that the result is a compile-time constant and can be used as such:

int my_array[increment<7>::result]; // array of 7+1 ints

Of course, defining an array like this is nonsense. Just as in run-time algorithms, incrementing is usually just one operation of an algorithm that computes something more complex.