Why case: always requires constant expression while if() doesn't?

Question

May be possible duplicate but couldn't have found the same.

Suppose I have following C code :

int a;
printf("Enter number :");
scanf("%d",&a);  // suppose entered only an integer
                // ignoring return value of scanf()

I got a case to check whether a is zero or non-zero.

if(a)
  printf("%d is non-zero",a);
else
  printf("%d is zero",a);

Everything is fine using if-else and I also know the other variations of if-else to achieve this . But problem comes with the switch-case as it says that we can implement everything in switch-case which we can do in if-else. But the following code fails.

switch(a)
{
 case a:
       printf("%d is non-zero",a);
       break;
 default:
       printf("%d is zero",a);
       break;
}

Also I know to reverse the case in the above code like this below will work and I will have my answer.

switch(a)
{
case 0:
    printf("%d is zero",a);
    break;
default :
    printf("%d is non-zero",a);
    break;
}

But the question is, Why ? Why if(a) is valid while case a: is not ? Is switch-case a compile time operation and if() run-time ?

Answer 1

The reason is that switch cases can be implemented as jump tables (typically using unconditional branch instructions). So they have to be resolved at compile time.

This makes them faster than ifs so it is better to use them when possible.