Kotlin位运算条件判断异常及IDE不可达警告问题排查

核心问题是Kotlin中位移运算符（shl）的优先级高于按位或运算符（or），导致代码的实际计算顺序和预期完全不同：

• 你预期的计算逻辑：每个条件对应的0/1先单独左移对应位数，再按位或合并，最终得到0b101（十进制5）：

  (1 shl 2) or (0 shl 1) or 1 = 4 or 0 or 1 = 5
  

• 实际计算顺序：由于shl优先级更高，代码会先计算(1 shl 2) or 0得到4，再将这个结果左移1位得到8，最后和1按位或得到0b1001（十进制9）。这个值不在你when分支的匹配列表中，因此会进入else分支；同时IDE会因为分支未覆盖该值，标记部分分支为不可达。

修复方案 #

给每个位移操作的结果单独加括号，强制指定计算顺序：

fun main(args: Array<String>) {
    val condition1 = true
    val condition2 = false
    val condition3 = true

    val state = ((if (condition1) 1 else 0) shl 2) or
            ((if (condition2) 1 else 0) shl 1) or
            (if (condition3) 1 else 0)

    println(state)

    when (state) {
        0b100 -> println("Only condition1 is true.")
        0b010 -> println("Only condition2 is true.")
        0b001 -> println("Only condition3 is true.")
        0b101 -> println("condition1 and condition3 are true.")
        0b110 -> println("condition1 and condition2 are true.")
        0b011 -> println("condition2 and condition3 are true.")
        0b111 -> println("All conditions are true.")
        else -> println("All conditions are false.")
    }
}

或者拆分变量，让逻辑更清晰：

fun main(args: Array<String>) {
    val condition1 = true
    val condition2 = false
    val condition3 = true

    val bitCondition1 = if (condition1) 1 shl 2 else 0
    val bitCondition2 = if (condition2) 1 shl 1 else 0
    val bitCondition3 = if (condition3) 1 else 0
    val state = bitCondition1 or bitCondition2 or bitCondition3

    println(state)

    when (state) {
        0b100 -> println("Only condition1 is true.")
        0b010 -> println("Only condition2 is true.")
        0b001 -> println("Only condition3 is true.")
        0b101 -> println("condition1 and condition3 are true.")
        0b110 -> println("condition1 and condition2 are true.")
        0b011 -> println("condition2 and condition3 are true.")
        0b111 -> println("All conditions are true.")
        else -> println("All conditions are false.")
    }
}

内容的提问来源于stack exchange，提问作者sho