Why is if (typeCar != 'X' || 'S' ||'L') is evaluating to always true

Question

cout << "Enter 'X' for uberX or 'S' for SUV or 'L' for luxury: "<< endl;
cin >> typeCar;
typeCar = toupper(typeCar);  
if (typeCar != 'X' || 'S' ||'L') 
{
    cout << "Enter X, S, or L"<<endl;
    system("pause"); 
}

Regardless of if I input 'X' 'S' or 'L', it always executes this statement. Same goes for when I put in lowercase.

`if (typeCar != 'X' || 'S' ||'L')` should be `if ((typeCar != 'X') && (typeCar != 'S') && (typeCar != 'L'))` if I understood correctly what you want to do. — Jesper Juhl, Nov 07 '17 at 08:19

OA617 · Answer 1 · 2017-11-07T08:22:31.507

-2

You need to evaluate each statement separately.

if(typechar! = 'x' || typechar! = 'Z')

edited Nov 07 '17 at 08:22

answered Nov 07 '17 at 08:19

OA617

76
6

if and not If also this answer is already written. – Tazo leladze Nov 07 '17 at 08:20
This is the auto correct in my phone... – OA617 Nov 07 '17 at 08:22

Why is if (typeCar != 'X' || 'S' ||'L') is evaluating to always true

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