Kotlin:必须初始化变量 'result'

Question

编译器显示错误 Kotlin: Variable result must be initialized.

这里是代码。

fun main(args: Array<String>) {
    print("Enter two numbers: ")

    // nextDouble() reads the next double from the keyboard
    var first= readLine()!!.toDouble()
    var second = readLine()!!.toInt()

    print("Enter an choice(1-4)): ")
    val operator = readLine()!!.toInt()

    var result: Double

    when (operator) {
        1 -> result = first + second
        2 -> result = first - second
        3 -> result = first * second
        4 -> result = first / second

        else -> {
            println("Error.")
        }
    }

    println("The result is :- " +result)
}

Answer 1

问题是当你在这里读取 result 的值时:

println("The result is :- " +result)

result 可能没有初始化，因为这里:

var result: Double

when (operator) {
    1 -> result = first + second
    2 -> result = first - second
    3 -> result = first * second
    4 -> result = first / second

    else -> {
        println("Error.")
    }

您没有为 else 分支中的 result 赋值。您有多种选择，一种可能是使 result 可以为空。例如:

var result = when (operator) {
    1 -> first + second
    2 -> first - second
    3 -> first * second
    4 -> first / second
    else -> null
}

if (result != null) {
    println("The result is :- " + result)
} else {
    println("Error.")
}

请注意，如果运算符不是 (1-4)，则 result 的值将为 null。此外，在您的代码中，您正在打印“错误”，然后您再次尝试打印结果。

要添加更多内容，如果您使用 when 语句作为方法引用来定义运算符，则可以使事情变得更好:

print("Enter an choice(1-4)): ")
val operatorCode = readLine()!!.toInt()

val operator: ((Int) -> Double)? = when (operatorCode) {
    1 -> first::plus
    2 -> first::minus
    3 -> first::times
    4 -> first::div
    else -> null
}

if (operator != null) {
    val result = operator.invoke(second)
    println("The result is :- " + result)
} else {
    println("Error.")
}