Swift:在某些闭包中无法分解元组(例如,使用枚举来减少)? [英] Swift: Unable to decompose tuple in certain closures (e.g., reduce with enumerate)?
问题描述
当使用map()和enumerate()时,Swift会分解枚举元组:
When using map() with enumerate(), Swift will decompose the enumerate tuple:
map(enumerate([1,2,3])) { (index, element) in
index + element
}
但是,这似乎不能与附加的闭包参数(例如,与reduce())一起工作:
However, this does not appear to work alongside an additional closure parameter (e.g., with reduce()):
reduce(enumerate([1,2,3]), 0) { (accum, (index, element)) in
accum + index + element
}
这会失败,并且错误:使用未声明的类型'index'
。
我缺少一些简单的东西,或者Swift简单地不允许分解一个元组和一个附加参数?我在1.1和1.2中尝试过。 (现在,我使用速记参数名称。)
Am I missing something simple, or does Swift simply not allow decomposing a tuple alongside an additional parameter? I have tried this in 1.1 and 1.2. (For now, I am using the shorthand argument names.)
推荐答案
答案是Swift根本不允许它。您不能像这样分解嵌套的元组。
The answer is "Swift simply not allow it". You can't decompose nested tuple like that.
例如,此代码编译,但我们不能访问 j
或 k
:
For example, this code compiles, but we cannot access j
or k
:
func foo(i: Int, (j: Int, k: Int) ) {
// println(j)
// ^ error: use of unresolved identifier 'j'
}
foo(1, (2, 3))
$ b >(j:Int,k:Int)只是一个类型,并且接收的元组本身没有名称。你必须这样写:
Because, in this code, (j: Int, k: Int)
is just a type, and the received tuple itself does not have a name. You have to write like this instead:
func foo(i: Int, x: (j: Int, k: Int) ) {
println(x.j)
}
foo(1, (1, 2))
$ b b
以同样的方式,这个编译,但它是无用的。
In the same way, this compiles, but it's useless.
reduce(enumerate([12,42,84]), 0) { (accum, (index:Int, element:Int)) in
相反,你必须接收元组本身,然后如果你想要分解它:
Instead, you have to receive the tuple itself, then decompose it if you want:
reduce(enumerate([12,42,84]), 0) { (accum, enumerated) in
println(enumerated.element)
let (index, element) = enumerated
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