can we write this in C++ switch?

#include <iostream>
using namespace std;

int main(){
ch开发者_运维技巧ar i;
cin >>i;
switch (i){
 case ('e'||'i'||'o'||'u'||'a'):
     cout<<"Vowel";
     break;
 case ('+'||'-'||'/'||'*'||'%'):
     cout<<"Op";
     break;
 }
return 0;  

}

if not than how can we use comparison or logical operators in switch ? & why cant we declare and initialize variable in single case without using scope ?

Without a break statement the previous cases "fall through" so this achieves the || you were looking for:

#include <iostream>
using namespace std;

int main(){
   char i;
   cin >>i;
   switch (i){
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'a':
        cout<<"Vowel";
        break;
   case '+':
   case '-':
   case '/':
   case '*':
   case '%':
        cout<<"Op";
        break;
   }
   return 0;  
}

The answer to the other part of your question is discussed in depth already on stackoverflow.

1. You can use fallthrough to map multiple case values to the same action.
2. The diagnostic message explains it -- it would be possible to jump over the initialization. But isn't it just a warning?

No, you can't; in switches you can only implicitly use the == operator and only on integral and enumeration types (§6.4.2). You should rewrite that switch as

switch (i){
 case 'e':
 case 'i':
 case 'o':
 case 'u':
 case 'a':
     cout<<"Vowel";
     break;
 case '+':
 case '-':
 case '/':
 case '*':
 case '%':
     cout<<"Op";
     break;
 }

which exploits the fall-through feature of the switch statement.

if not than how can we use comparison or logical operators in switch ?

Simply, you can't. If you want to do anything different than equality comparison with integral/enumeration types you have to write several if/else statements.

& why cant we declare and initialize variable in single case without using scope ?

It's not a problem of declaration, but of initialization; see the link in @awoodland's answer.

Format it like this:

switch (i)
{
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
        cout << "Vowel";
        break;
}

Alternative, more terse solution:

#include <cstring>

// ...

if (strchr("eioua", i)) cout << "vowel";
if (strchr("+-/*%", i)) cout << "operator";

Note that strchr considers the terminating zero part of the string, so i should not be 0.

We could, except it doesn't mean what is intended (and yields the same value in both cases): you'd be performing a logical or on a bunch of non-zero integer values and the result is true in both cases.