if integer is greater than x but less than y (Swift)

Question

I was wondering if there is any smooth way of checking the value of an integer in a range in swift.

I have an integer that can be any number between 0 and 1000

I want to write an if statement for "if the integer is between 300 and 700 - do this and if its any other number - do something else"

I could write:

if integer > 300 {
    if integer < 700 {
      //do something else1
    }
    //do something
} else {
    // do something else2

}

But I want to minimize the amount of code to write since "do something else1" and "do something else2" are supposed to be the same

It doesn't seem that you can write :

if 300 < integer < 700 {

} else {

}

I tried using

if integer == 300..<700 {
}

but that didn't work either. Anybody got a suggestion?

Answer 1

Did you try this?

if integer > 300 && integer < 700 {
    // do something
} else {
    // do something else               
}

Hope this helps

Answer 2

There is a type, HalfOpenInterval, that can be constructed with ... and that has a .contains method:

if (301..<700).contains(integer) {
    // etc.
}

Note, 301..<700 on its own will create a Range, which doesn’t have a contains function, but Swift’s type inference will see that you’re calling a method that only HalfOpenInterval has, and so picks that particular overload.

I mention this just because if you want to store the interval as a variable instead of using it in-line, you need to specify:

let interval = 301..<700 as HalfOpenInterval
if interval.contains(integer) {
  // etc.
}