问题
So, I have a for-loop that looks similar to this:
for var i = 0; i < results.count ; i += 1 {
if (results[i] < 5) {
results.removeAtIndex(i)
i -= 1
}
}
This used to work. But when I changed it to the preferred Swift 3.0 syntax:
for var i in 0..<results.count {
if (results[i] < 5) {
results.removeAtIndex(i)
i -= 1
}
}
I get an array IOOBE exception because it doesn't re-check the count and continues on until the original results.count.
How do I fix this? It works now, but I don't want to get into trouble in the future.
回答1:
While the solution making use of filter is a fine solution and it's more Swift-ly, there is another way, if making use of for-in is, nonetheless, still desired:
var results = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for var i in (0..<results.count).reverse()
{
if (results[i] < 5)
{
results.removeAtIndex(i)
//i -= 1
}
}
print(results)
Result:
[5, 6, 7, 8, 9, 10]
We could omit this line i -= 1 altogether, in addition.
The problem with removeAtIndex within the loop is that it will not cause the array to re-index itself in-place and thus causing an array out of bounds exception due to count not being updated.
By traversing backwards, the out of bounds exception can thus be avoided.
回答2:
Could you use a filter instead?
let numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let greaterThan5 = numbers.filter{$0 >= 5}
print(greaterThan5)
回答3:
If you want to continue using a for-loop, you can enumerate over both index and element using enumerate:
for (index, element) in results.enumerate() {
if (element < 5) {
results.removeAtIndex(index)
}
}
Although depending on what you're doing in your loop, the filter method might be a better idea.
回答4:
If your loop goes forward...
for var i in (0..<results.count) where results.indices.contains(i) {
//if the index doesn't exist, the loop will be stopped.
if (results[i] < 5) {
results.removeAtIndex(i)
}
}
来源:https://stackoverflow.com/questions/37596063/how-do-i-write-a-for-loop-in-swift-3-for-an-array-that-i-modify-during-the-for-l