Why can switch only compare to const values?

Question

Just curious:

Take a switch-statement like -

switch(myVar)
{
    case(label):
    ...
    break;
}

why label should be constant only?

Answer 1

Because it can be ambiguous otherwise:

int a = 1, b = 1;
int val = 1;

switch (val) {
case a: foo(); break;
case b: bar(); break;
}

Requiring constant expressions for the cases makes sure that there's
no ambiguity, since they can be evaluated at compile time.

Answer 2

Because thats how its defined in The Standard.
Use if() if you want to compare variables.

Answer 3

How would you be able to define behavior by comparing to a variable
that ranges over a large set of values? How do you limit what values
the variable can have? It wouldnt even make sense to compare to a
variable, you are defining the behavior in the separate statements in
a switch based on a specific state, it amounts to using an if
statement... if myVar=1, do this, if myVar=2, do this, and so on. If
you need to use a range of values, you would need to use an explicit
if: if myVar > 0 && myVar < 100, do this, if myVar > 100 & If I were
to divulge it, it would overturn the world."

Comments

Compiler will be optimised for switch statement. If there are case
from 1 to 100 and the value of switching variable is 76, then the code will
not run like compare case 1, case 2, case 3,... Case 76. The code will be
optimised to reach case 76 much more faster by using some search
methods.
This optimisation can only happen if we have a sequential variable
which is obviously integers.

---

Theres no searching on most common platforms. An optimizing compiler
 will build a "jump table" (a list of addresses) and simply branch to
the  target code based on the index specified in the jump table.

 But I posit that if you have 76 case statements in a single switch
 block, youre probably doing something wrong and need to refactor your

 code. I realize you meant the 76 as an example though because youve
 never written any code like that.