Swift compiler optimization of 'for' loops

Question

I have a 'for' loop similar to this:

for i in 0...n - 1 {
    // do stuff, n is not changed
}

Back in the day, when I coded in FORTRAN, I was taught that this was a very inefficient way to code large 'for' loops.

Does the Swift compiler recognize the non-changing limit and pre-calculate n - 1?

It is quite easy to find out: Replace n by some function n() and check how often the function is called ... — Martin R
– Martin R, Commented Mar 20, 2016 at 9:28

Sweeper · Accepted Answer · 2016-03-20 09:37:35Z

2

I think the answer is yes.

1...n - 1 represents a Range object. It is a literal of Range. Therefore, when compilation reaches the loop and sees the literal, it thinks

It seems like that you want to create a new Range<Int> object! Here you go! Hmm... So I guess i is of type Int...

and so on.

This means that n - 1 is evaluated when you create the object. And it stays that way, not evaluating it a second time. This code proves it by not printing only one hello:

var n = 10
for i in 1...n - 1 {
    n = 2
    print("Hello")
}

So yeah.

Note:

It's better to use 1..<n instead of 1...n - 1 in this case, they are the same.
Apple actually recommends you to use this approach instead of the C-style for loop.
The C-style for loop will be removed in Swift 3

edited Mar 20, 2016 at 9:37

answered Mar 20, 2016 at 9:28

Sweeper

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5 Comments

Martin R Over a year ago

It is better to use 0..<n, but why?

Sweeper Over a year ago

It's my opinion... More swiftier, uses fewer characters, and not unclear. @MartinR

Martin R Over a year ago

If n == 0 then 0 ... n-1 aborts with a runtime exception, but 0 ..< n doesn't (and executes the loop body zero times).

Martin R Over a year ago

Some nitpicking: 1...n - 1 is not a literal. It is the result of the ... operator applied to the integer literal 1 and the expression n-1.

Simon Youens Over a year ago

Thanks all. I like the <n approach.

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