String 如何在swift 3.0中连接多个可选字符串?
我试图在swift 3中连接多个字符串:String 如何在swift 3.0中连接多个可选字符串?,string,swift,swift3,String,Swift,Swift3,我试图在swift 3中连接多个字符串: var a:String? = "a" var b:String? = "b" var c:String? = "c" var d:String? = a! + b! + c! 编译时出现以下错误: error: cannot convert value of type 'String' to specified type 'String?' var d:String? = a! + b! + c! ~~~~~~~~^~
var a:String? = "a"
var b:String? = "b"
var c:String? = "c"
var d:String? = a! + b! + c!
编译时出现以下错误:
error: cannot convert value of type 'String' to specified type 'String?'
var d:String? = a! + b! + c!
~~~~~~~~^~~~
这曾经在swift 2中起作用。我不知道为什么它不再工作。OP提交的错误报告:
这似乎是一个与以下情况相关的bug(Swift 2.2中不存在,仅3.0中存在):
- 对表达式中的至少3个术语使用强制展开运算符(
)(使用至少2个基本运算符测试,例如!
或+
)-
- 出于某种原因,鉴于上述情况,Swift会弄乱表达式的类型推断(特别是对于表达式中的
术语本身)x!
let a: String? = "a"
let b: String? = "b"
let c: String? = "c"
存在错误:
// example 1
a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 2
var d: String = a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 3
var d: String? = a! + b! + c!
/* error: cannot convert value of type 'String'
to specified type 'String?' */
// example 4
var d: String?
d = a! + b! + c!
/* error: cannot assign value of type 'String'
to specified type 'String?' */
// example 5 (not just for type String and '+' operator)
let a: Int? = 1
let b: Int? = 2
let c: Int? = 3
var d: Int? = a! + b! + c!
/* error: cannot convert value of type 'Int'
to specified type 'Int?' */
var e: Int? = a! - b! - c! // same error
/* example 1 */
var d: String? = a! + b!
/* example 2 */
let aa = a!
let bb = b!
let cc = c!
var d: String? = aa + bb + cc
var e: String = aa + bb + cc
/* example 3 */
var d: String? = String(a!) + String(b!) + String(c!)
错误不存在:
// example 1
a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 2
var d: String = a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 3
var d: String? = a! + b! + c!
/* error: cannot convert value of type 'String'
to specified type 'String?' */
// example 4
var d: String?
d = a! + b! + c!
/* error: cannot assign value of type 'String'
to specified type 'String?' */
// example 5 (not just for type String and '+' operator)
let a: Int? = 1
let b: Int? = 2
let c: Int? = 3
var d: Int? = a! + b! + c!
/* error: cannot convert value of type 'Int'
to specified type 'Int?' */
var e: Int? = a! - b! - c! // same error
/* example 1 */
var d: String? = a! + b!
/* example 2 */
let aa = a!
let bb = b!
let cc = c!
var d: String? = aa + bb + cc
var e: String = aa + bb + cc
/* example 3 */
var d: String? = String(a!) + String(b!) + String(c!)
然而,由于这是Swift 3.0-dev,我不确定这是否真的是一个“bug”,以及w.r.t.在尚未生产的代码版本中报告“bug”的策略是什么,但可能您应该为此提交雷达,以防万一 关于如何规避这一问题,请回答您的问题:
- 例如,使用上述示例2中未出现的错误中的中间变量
- 或者明确地告诉Swift,3项表达式中的所有术语都是字符串,如上面的示例3中的Bug not present所示
- 或者,更好的方法是使用可选的安全展开,例如使用可选绑定:
compactMap可用于从展平数组中过滤出nil值谢谢你的回答,我在你的代码可能运行时提交了一个bug,如果你能在回答中添加一些解释就好了。
let q: String? = "Hello"
let w: String? = "World"
let r: String? = "!"
var array = [q, w, r]
print(array.flatMap { $0 }.reduce("", {$0 + $1}))
// HelloWorld!
let q: String? = "Hello"
let w: String? = nil
let r: String? = "!"
var array = [q, w, r]
print(array.flatMap { $0 }.reduce("", {$0 + $1}))
// Hello!
func getSingleValue(_ value: String?..., seperator: String = " ") -> String? {
return value.reduce("") {
($0) + seperator + ($1 ?? "")
}.trimmingCharacters(in: CharacterSet(charactersIn: seperator) )
}
var a:String? = "a"
var b:String? = "b"
var c:String? = "c"
var d:String? = ""
let arr = [a,b,c]
arr.compactMap { $0 }.joined(separator: " ")