why I get Different result in switch statements

Question

I am learning JavaScript and currently I build simple tip calculator using function and switch statement. Basically, tip 20% of the bill when the bill is less than $50, 15%

Answer 1

When you do

switch(bill){

a case will be fulfilled if the expression that follows it is === to the value of bill. For example, if bill is 124, then switch(bill) would require

case: 124:
    tip = bill * .15 // because 124 <= 200

for your program to work as expected. Your code is producing unexpected results because all the cases fail, and it falls through to the default.

The switch(true) works because when you have cases of

case bill > 0 && bill < 50:

this will effectively evaluate to

case true
// block below will run, because `true` is `===` to the expression that was switched against

or

case false:
// block below will not run, because `false` is `!==` to the expression that was switched against

and run the case block accordingly.

switch is not the right tool for the job, IMO - it's confusing and verbose. (It's never the right tool for any programming task in Javascript, I think.) I'd use if/else instead:

const getTip = (bill) => {
  if (bill > 0 && bill < 50) {
    return bill * 0.2;
  } else if (bill >= 50 && bill <= 200) {
    return bill * 0.15;
  } else {
    return bill * 0.1;
  }
};

function simpleTipCalculator(bill) {
  console.log(getTip(bill));
}

simpleTipCalculator(124)
simpleTipCalculator(48)
simpleTipCalculator(268)