How to specify a range to avoid in Swift

Question

I'm trying to make it so when I provide a value that isn't in the thirties it prints out a certain message. Here is my code:

let age = 25

if case 18...25 = age {
  print("Cool demographic")
}
else if case !(30...39) = age {
  print("not in there thirties")
}

Leo Dabus · Answer 1 · 2018-12-22T19:34:59.880

You can use pattern match operator ~=

static func ~= (pattern: Range<Bound>, value: Bound) -> Bool

You can use this pattern matching operator (~=) to test whether a value is included in a range. The following example uses the ~= operator to test whether an integer is included in a range of numbers.

let age = 29

if 18...25 ~= age {
    print("Cool demographic")
} else if !(30...39 ~= age) {
    print("not in there thirties") // "not in there thirties\n"
}

score 2 · Answer 2 · answered Dec 10 '18 at 02:41

I like contains rather than the unnecessarily obscure if case. There are situations where if case is needed, but this is not one of them. It's good to say what you mean. So:

let age = 25

if (18...25).contains(age) {
    print("Cool demographic")
}
else if !(30...39).contains(age) {
    print("not in their thirties")
}

score 0 · Answer 3 · answered Dec 09 '18 at 23:54

0

If age is an integer, you can do a direct comparison:

if age < 30 || age > 39 {
    print("not in thirties")
}

answered Dec 09 '18 at 23:54

Gniem

111
4

Thank you that worked ! Why didn't I think of that :P – Aezharhu Dec 09 '18 at 23:55

score 0 · Answer 4 · answered Dec 10 '18 at 14:08

You can also use switch, in my opinion it's a little bit more expressive than if-else:

let age = 25

switch age {
case 18...25:
    print("Cool demographic")
case _ where !(30...39 ~= age):
    print("not in there thirties")
default:
    break
}

You can find good examples with switch from Imanou Petit by this link:

How to specify a range to avoid in Swift

4 Answers4