[swift-evolution] A (better) Swift Equivalent For The Classical For-Loop With Numeric Scalars

[Dany St-Amant]

> Le 22 mars 2016 à 21:19, William Dillon <william at housedillon.com> a écrit :
> 
>> 
>> These may be compact, but some programmer may fell that they are not in control: do these "stdlib" seq(), c.filter() pre-calculate all the entries (wasting precious cpu cycle if one break out early)  or are each value dynamically created? Having an explicit legacy loop format gives a feel of control. So something like
>> 
>> for i from 2 to 1_000_000 by 1 where i % 2 != 0 while foundCount < 5 { print(i); foundCount +=1 }
>> 
>> sounds less magical and seems easier to predict than
>> 
>> for i in 2.stride(to:1_000_000, by:1).filter({ $0 % 2 != 0}).prefix(5) { print(i) }
>> 
>> and is still quite readable, even if it mixes for loop, while loop and even simple condition.
>> 
> 
> I disagree.  I think the first case is venturing dangerously close to the AppleScript “uncanny valley” of human language-like programming verbs where I never know the exact words and exact order for things.  The second example is right out of any functional programming mold, and I fully understand what’s happening, and how to decompose the process to test assumptions.  Also, I know how to implement the second case (more or less) from scratch, and that’s pretty huge for me.
> 

There have been multiple fights over too verbose versus too terse, and on whether or not making it sound like English. I would agree a verbose syntax for a mandatory statement is a pain, but the while and where above are optional. And using them instead of relying on continue or break allow for code intent to be bit clearer.

Reusing my bad example, without the from/to/by:

for i in 2.stride(to: 1_000_000, by: 1)
{
	if i % 2 != 0 { continue }
	print(i)
	foundCount += 1
	if foundCount >= 5 { break }
}

// Where clause already work in current Swift
for i in 2.stride(to: 1_000_000, by: 1)
where i % 2 == 0
{
	print(i)
	foundCount += 1
	if foundCount >= 5 { break }
}

// While do not exist yet in this context
for i in 2.stride(to: 1_000_000, by: 1)
where i % 2 == 0
while foundCount < 5
{
	print(i)
	foundCount += 1
}

Dany


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