Swift: how to perform a task n-1 times where n is the length of an array? -

i came this:

for x in 1...(myarray.count - 1) { task() } 

which ugly. there better way?

you have little careful, if array empty, crash:

let a: [int] = [] let range = 0..<a.count-1 // fatal error: can't form range end < start 

strides don’t have problem (since strideable things must comparable) do:

for _ in stride(from: 0, to: a.count - 1, by: 1) {     // execute if a.count > 2     print("blah") } 

alternatively, if guard it, can use dropfirst:

for _ in (a.isempty ? [] : dropfirst(a))  {     print("blah") } 

i advise against trying make neater creating pseudo-for-loop runs 1 less count times. there’s reason there’s no foreach or repeat functions in swift. these kind of loops seem nice @ first there lots of bugs arise them (for example, return or continue don’t work way might expect, it’s considered bad practice use higher-order function external mutation – whereas regular for loop suggestion mutation/side-effects likely).

the neatest extension-type solution extend array safe drop-first:

extension array {     // version of dropfirst returns empty array both     // empty , single-element array     func safedropfirst() -> arrayslice<t> {         return self.isempty ? [] : dropfirst(self)     } }  _ in myarray.safedropfirst() {     dothing() }