Optimized way to check the value of a variable with enum members value

Question

Hello I have the code below:

enum {a, b, c, d, ..., z} abc;

int main()
{
  int val = 20;
  if (val == a || val == b ||val == c||val == d..... || val == z)
  {
    /*Do something*/
  }
}

Is there any other way so that we can skip the OR operation because if there are 1000s of enum members then how can we do ahead with checking with all members. Please help.

Answer 1

A modern compiler should just be able to optimize such code if, as in your case, the value of the expression is known at compile time. For readability and error checking I think that using a switch would be better:

switch (val)  {
 case a:;
 case b:;
 ....
 // your code goes here
}

As said, performance wise there shouldn't be much difference, the compiler will transform this to a table lookup (or other clever things) if appropriate or completely optimize it out if val is known at compile time.

But you can have the advantage of error checking compilers, here. If you don't have a default case, most compilers will warn you if you omit one of the enumeration constants. Also I think that this is clearer, since it doesn't repeat the evaluation of val all over the place.

Answer 2

other(faster) solution will be the following

bool isInenum (int val)
{
    bool retVal = false
        switch(val)

    {
        case a:
        case b:
        case c:
        case d:
            {
                retVal = true;
            }

    }
    return retVal;
}