xcode 错误：无法使用 Swift + PromiseKit 将类型 '() -> ()' 的值转换为闭包结果类型 'String'
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error: cannot convert value of type '() -> ()' to closure result type 'String' using Swift + PromiseKit
提问by John Mike
I am new to promises in Swift and using PromiseKit to try and create a very simple response in a playground and attempt to use it. I have the following code:
我是 Swift 中的 Promise 的新手，并使用 PromiseKit 在操场上尝试创建一个非常简单的响应并尝试使用它。我有以下代码：
//: Playground - noun: a place where people can play
import UIKit
import PromiseKit
func foo(_ error: Bool) -> Promise<String> {
return Promise { fulfill, reject in
if (!error) {
fulfill("foo")
} else {
reject(Error(domain:"", code:1, userInfo:nil))
}
}
}
foo(true).then { response -> String in {
print(response)
}
}
However I get the following error:
但是我收到以下错误：
error: MyPlayground.playground:11:40: error: cannot convert value of type '() -> ()' to closure result type 'String'
foo(true).then { response -> String in {
error: MyPlayground.playground:11:40: error: cannot convert value of type '() -> ()' to closure result type 'String'
foo(true).then { response -> String in {
采纳答案by aaplmath
The error is being thrown because the closure you're passing to thenpurports to return a String, but no such value is ever returned. Unless you plan on returning a Stringsomewhere in that closure, you need to change the closure's return type to Void, as follows:
抛出错误是因为您传递给的闭包then声称要返回 a String，但从未返回过这样的值。除非您打算String在该闭包中的某处返回 a，否则您需要将闭包的返回类型更改为Void，如下所示：
foo(true).then { response -> Void in
print(response)
}
Note that closures with the return type Voidcan have their return type omitted. Also, you have an extraneous {in the code in your question (I'm assuming this doesn't exist in your actual code, since it compiles).
请注意，具有返回类型的闭包Void可以省略其返回类型。此外，您{的问题中的代码中有一个无关紧要的内容（我假设这在您的实际代码中不存在，因为它可以编译）。
In addition to that issue, Errorhas no accessible initializers—the initializer you're using in your code actually belongs to NSError, and thus your rejectcall needs to look like:
除了这个问题，Error没有可访问的初始化程序——您在代码中使用的初始化程序实际上属于NSError，因此您的reject调用需要如下所示：
reject(NSError(domain:"", code:1, userInfo:nil))
