Question

我正在尝试将Int16值转换为String，但它使用Optional赋值，并且不允许我强行打开它。

String(describing:intValue)

结果：可选（intValue）

Answer 1

先打开intValue，然后将其传递给字符串初始值设定项：String(unwrappedIntValue)

以下是处理可选项的一些方法。我已经添加了带有类型注释的显式string#变量，以明确涉及的类型

let optionalInt: Int? = 1

// Example 1
// some case: print the value within `optionalInt` as a String
// nil case:  "optionalInt was nil"

if let int = optionalInt {
    let string1: String = String(int)
    print(string1)
}
else {
    print("optionalInt was nil")
}

// Example 2, use the nil-coalescing operator (??) to provide a default value
// some case: print the value within `optionalInt` as a String
// nil case:  print the default value, 123

let string2: String = String(optionalInt ?? 123)
print(string2)

// Example 3, use Optional.map to convert optionalInt to a String only when there is a value
// some case: print the value within `optionalInt` as a String
// nil case:  print `nil`

let string3: String? = optionalInt.map(String.init)
print(string3 as Any)

// Optionally, combine it with the nil-coalescing operator (??) to provide a default string value
// for when the map function encounters nil:
// some case: print the value within `optionalInt` as a String
// nil case:  print the default string value "optionalInt was nil"

let string4: String = optionalInt.map(String.init) ?? "optionalInt was nil"
print(string4)

Answer 2

您可以使用字符串插值将数字转换为String：

let stringValue = "\(intValue)"

或者您可以使用标准String初始化程序：

let stringValue = String(intValue)

如果数字是Optional，请先打开它：

let optionalNumber: Int? = 15

if let unwrappedNumber = optionalNumber {
  let stringValue = "\(unwrappedNumber)"
}

或者

if let unwrappedNumber = optionalNumber {
  let stringValue = String(unwrappedNumber)
}

将Int16转换为String Swift 4

2 个答案: