why does this simple switch statement always run the default

Question

Ok guys, It seems like this switch statement is forever doomed to NOT work.

The initial idea was to create a variable x which is a prompt, the user will have to select enter any number and that would be the value of x.

Then under the first case of the switch, if x is less than 0.5 then it will simply console.log "less". If x is more than 0.5 it will simply console.log "more". If for some reason the program didn't work as expected the default is to console.log "this is the default"

Then i added a console.log of x in the end just to know what number did the user enter.

Lets try it!

I tried and tried and regardless of what number i enter it always printed "this is the default". Then printed the value of x.

I ended up going Rambo and removing the prompt and declaring x to be 0.6. It ought to print "more" but it still doesn't.

var x = 0.6;

switch (x) {
  case x < 0.5:
    console.log("less");
    break;
  case x > 0.5:
    console.log("more");
    break;

  default:
    console.log("its the dflt");
};

console.log(x);

So I'm wondering whats wrong with this code. Help

Thanks certainperformance, i was sure someone was bewitching my code lol — Cool_berserker, Jun 16 '19 at 07:24

CertainPerformance · Accepted Answer · 2019-06-15T22:08:42.080

switch compares what you switch with against the cases. So, if you have case x < 0.5: which you want to run, that case will run if the expression you switched against was true:

var x = 0.6;

switch (true) {
  case x < 0.5:
    console.log("less");
    break;
  case x > 0.5:
    console.log("more");
    break;

  default:
    console.log("its the dflt");
};

console.log(x);

If you switch against x itself, a case will only run if the case evaluates to the same value as x, which, here, is 0.6, eg:

var x = 0.6;

switch (x) {
  case 0.6:
    console.log('x is exactly 0.6');
    break;
  default:
    console.log("x is something other than 0.6");
};

console.log(x);

But that's not flexible at all, and isn't what you want.

Personally, I'd prefer if/else, it's a lot easier to read (and, as some points out in comments, is a lot faster):

var x = 0.6;
if (x < 0.5) {
    console.log("less");
} else if (x > 0.5) {
    console.log("more");
} else {
    console.log('neither less nor more; equal or NaN');
}

Or `switch (x>5)`, `case true/false` and `default`, although using a `switch` for OP's use case is rather redundant anyway. — Yannick K, Jun 15 '19 at 21:59
Please do not use `switch(true)` https://stackoverflow.com/a/12259830/36866 — some, Jun 15 '19 at 22:04

score 0 · Answer 2 · answered Jun 15 '19 at 21:58

Switch compares the value of x to the value of the cases. In your code x < 0.5 evaluates to true. Instead of going to that case like if-statements, the switch case compares x and true. Since x is a number, x will never equal true so the default case is always taken.

I would use if-statements instead of a switch in this instance. Switches are better for enumerations (checking if x is a specific value out of a set of values, not a range of values)

score 0 · Answer 3 · answered Jun 15 '19 at 22:04

CertainPerformance has answered you question very well however if you still don't understand how to use switch I would recommend you use "if statements" until you have the time to read more on using switch.

var x = 0.6;

if (x < 0.5) {
  console.log("less");
}
else if (x > 0.5) {
  console.log("more");
}
else {
  console.log("its the dflt");
}

console.log(x);

Hope this is easier for you :)

why does this simple switch statement always run the default

3 Answers3