How to correctly write a condition in switch?

Question

Good afternoon! Why does the first option work - switch (true), and the second option does not work - switch (a)?

First:

var a= prompt('Enter value', '');
switch(true)
{
    case a>10:
        alert('a>10');
    break;

    case a<10:
        alert('a<10');
    break;

    default:
        alert('a===10');

Second:

var a= prompt('Enter value', '');
switch(a)
{
    case a>10:
        alert('a>10');
    break;

    case a<10:
        alert('a<10');
    break;

    default:
        alert('a===10');

That's an interesting use of `switch`. – Derek 朕會功夫 Nov 06 '17 at 07:03 — Derek 朕會功夫, Nov 06 '17 at 07:03

score 4 · Accepted Answer · answered Nov 06 '17 at 07:04

4

Why does the first option work - switch (true), and the second option does not work - switch (a)?

As per documentation

The switch statement evaluates an expression, matching the expression's value to a case clause, and executes statements associated with that case.

So, in your first option true will match to either a < 10 or a > 10, however in second option, a being a string may not match to either of them.

answered Nov 06 '17 at 07:04

gurvinder372

66,980
10
72
94

1

at last proper answer for question. – Munkhdelger Tumenbayar Nov 06 '17 at 07:06
Спасибо! Сейчас it's нормально – Alexey_Danilov Nov 06 '17 at 07:16
maybe worth to mention the strict comparison. – Nina Scholz Nov 06 '17 at 07:18

AngYC · Answer 2 · 2017-11-06T07:16:13.537

2

Edit: I just realize OP ask for the difference instead of why it won't work, sorry for misunderstanding the question

It should work nicely

var a = prompt('Enter value', '');

switch (true) {
    case (a > 10):
        alert("a > 10");
        break;

    case (a < 10):
        alert("a < 10");
        break;

    default:
        alert('a == 10');
}

edited Nov 06 '17 at 07:16

answered Nov 06 '17 at 07:03

AngYC

3,051
6
20

StackSlave · Answer 3 · 2017-11-06T07:20:51.623

It's because a > 10 is true, like the switch(true), while switch(a) was passed a, which is not true. Of course, you should cast a. a = +a or use parseInt() or parseFloat().

Here's what you probably meant to do:

var a = prompt('Enter value');
if(+a > 10){
  alert('a > 10');
}
else if(a !== '' && +a < 10){
  alert('a < 10');
}
else if(+a === 10){
  alert('a === 10'); 
}
else{
  alert('Really, you should avoid using prompt and alert!');
}
// notice this is less code than that pointless switch

score -2 · Answer 4 · answered Nov 06 '17 at 07:02

-2

You need to convert the user input from a string to an integer, like so

a = parseInt(a)

answered Nov 06 '17 at 07:02

newman

424
2
12

JS already does it for you. It's because of the `switch(a)` part. – Derek 朕會功夫 Nov 06 '17 at 07:04

How to correctly write a condition in switch?

4 Answers4