Swift 中的反向范围 [英] Reverse Range in Swift
问题描述
有没有办法在 Swift 中使用反向范围?
例如:
for i in 5...1 {//做点什么}
是一个无限循环.
在较新版本的 Swift 中,代码可以编译,但在运行时出现错误:
<块引用>致命错误:无法形成带有 upperBound
我知道我可以使用 1..5
代替,计算 j = 6 - i
并使用 j
作为我的索引.我只是想知道有没有更清晰的东西?
更新最新的 Swift 3(仍然适用于 Swift 4)
您可以在范围上使用 reversed()
方法
for i in (1...5).reversed() { print(i) }//5 4 3 2 1
或者stride(from:through:by:)
方法
for i in stride(from:5,through:1,by:-1) { print(i) }//5 4 3 2 1
stide(from:to:by:)
类似但不包括最后一个值
for i in stride(from:5,to:0,by:-1) { print(i) }//5 4 3 2 1
更新最新的 Swift 2
首先,协议扩展改变了reverse
的使用方式:
for i in (1...5).reverse() { print(i) }//5 4 3 2 1
Stride 已在 Xcode 7 Beta 6 中重新设计.新用法是:
for i in 0.stride(to: -8, by: -2) { print(i) }//0 -2 -4 -6for i in 0.stride(through: -8, by: -2) { print(i) }//0 -2 -4 -6 -8
它也适用于双打
:
for i in 0.5.stride(to:-0.1, by: -0.1) { print(i) }
小心浮点数比较这里的界限.
Swift 1.2 的早期编辑:从 Xcode 6 Beta 4 开始,by 和 ReverseRange 不再存在:[
如果您只是想反转一个范围,reverse 函数就是您所需要的:
for i in reverse(1...5) { println(i) }//打印 5,4,3,2,1
正如 0x7fffffff 发布的那样,有一个新的 stride 构造,可用于迭代和递增任意整数.Apple 还表示即将提供浮点支持.
来自他的回答:
for x in stride(from: 0, through: -8, by: -2) {println(x)//0, -2, -4, -6, -8}对于 x 步幅(从:6,到:-2,由:-4){println(x)//6, 2}
Is there a way to work with reverse ranges in Swift?
For example:
for i in 5...1 {
// do something
}
is an infinite loop.
In newer versions of Swift that code compiles, but at runtime gives the error:
Fatal error: Can't form Range with upperBound < lowerBound
I know I can use 1..5
instead, calculate j = 6 - i
and use j
as my index. I was just wondering if there was anything more legible?
Update For latest Swift 3 (still works in Swift 4)
You can use the reversed()
method on a range
for i in (1...5).reversed() { print(i) } // 5 4 3 2 1
Or stride(from:through:by:)
method
for i in stride(from:5,through:1,by:-1) { print(i) } // 5 4 3 2 1
stide(from:to:by:)
is similar but excludes the last value
for i in stride(from:5,to:0,by:-1) { print(i) } // 5 4 3 2 1
Update For latest Swift 2
First of all, protocol extensions change how reverse
is used:
for i in (1...5).reverse() { print(i) } // 5 4 3 2 1
Stride has been reworked in Xcode 7 Beta 6. The new usage is:
for i in 0.stride(to: -8, by: -2) { print(i) } // 0 -2 -4 -6
for i in 0.stride(through: -8, by: -2) { print(i) } // 0 -2 -4 -6 -8
It also works for Doubles
:
for i in 0.5.stride(to:-0.1, by: -0.1) { print(i) }
Be wary of floating point compares here for the bounds.
Earlier edit for Swift 1.2: As of Xcode 6 Beta 4, by and ReverseRange don't exist anymore :[
If you are just looking to reverse a range, the reverse function is all you need:
for i in reverse(1...5) { println(i) } // prints 5,4,3,2,1
As posted by 0x7fffffff there is a new stride construct which can be used to iterate and increment by arbitrary integers. Apple also stated that floating point support is coming.
Sourced from his answer:
for x in stride(from: 0, through: -8, by: -2) {
println(x) // 0, -2, -4, -6, -8
}
for x in stride(from: 6, to: -2, by: -4) {
println(x) // 6, 2
}
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