How to make a for loop variable const with the exception of the increment statement?

Question

Consider a standard for loop:

for (int i = 0; i < 10; ++i) 
{
   // do something with i
}

I want to prevent the variable i fr

Answer 1

#include <cstdio>
  
#define protect(var) \
  auto &var ## _ref = var; \
  const auto &var = var ## _ref

int main()
{
  for (int i = 0; i < 10; ++i) 
  {
    {
      protect(i);
      // do something with i
      //
      printf("%d\n", i);
      i = 42; // error!! remove this and it compiles.
    }
  }
}

---

Note: we need to nest the scope because of an astonishing stupidity in the language: the variable declared in the for(...) header is considered to be at the same nesting level as variables declared in the {...} compound statement. This means that, for instance:

for (int i = ...)
{
  int i = 42; // error: i redeclared in same scope
}

What? Didn't we just open a curly brace? Moreover, it's inconsistent:

void fun(int i)
{
  int i = 42; // OK
}

Answer 2

Couldn't you just move some or all the content of your for loop in a function that accepts i as a const?

Its less optimal than some solutions proposed, but if possible this is quite simple to do.

Edit: Just an example as I tend to be unclear.

for (int i = 0; i < 10; ++i) 
{
   looper( i );
}

void looper ( const int v )
{
    // do your thing here
}

Answer 3

The KISS version...

for (int _i = 0; _i < 10; ++_i) {
    const int i = _i;

    // use i here
}

If your use case is just to prevent accidental modification of the loop index then this should make such a bug obvious. (If you want to prevent intentional modification, well, good luck...)