您可以在Kotlin中以多种方式实现它

如果使用 如果(a) b else c 使用时 当(a) { True ->打印("value b") 错误->打印("value c") Else -> { 打印("在任何其他情况下默认返回") ｝ ｝ 空安全 Val a = b ?: c

例如:var energy: Int = data?.get(position)?.energy?.toInt() ?: 0

在kotlin中，如果你使用?:，它会像这样工作，如果语句将返回null，那么?:0，它将取0或任何你写在这一边的东西。

使用以下中缀函数，我可以覆盖许多常见的用例，几乎与在Python中实现的方式相同:

class TestKotlinTernaryConditionalOperator {

    @Test
    fun testAndOrInfixFunctions() {
        Assertions.assertThat(true and "yes" or "no").isEqualTo("yes")
        Assertions.assertThat(false and "yes" or "no").isEqualTo("no")

        Assertions.assertThat("A" and "yes" or "no").isEqualTo("yes")
        Assertions.assertThat("" and "yes" or "no").isEqualTo("no")

        Assertions.assertThat(1 and "yes" or "no").isEqualTo("yes")
        Assertions.assertThat(0 and "yes" or "no").isEqualTo("no")

        Assertions.assertThat(Date() and "yes" or "no").isEqualTo("yes")
        @Suppress("CAST_NEVER_SUCCEEDS")
        Assertions.assertThat(null as Date? and "yes" or "no").isEqualTo("no")
    }
}

infix fun <E> Boolean?.and(other: E?): E? = if (this == true) other else null
infix fun <E> CharSequence?.and(other: E?): E? = if (!(this ?: "").isEmpty()) other else null
infix fun <E> Number?.and(other: E?): E? = if (this?.toInt() ?: 0 != 0) other else null
infix fun <E> Any?.and(other: E?): E? = if (this != null) other else null
infix fun <E> E?.or(other: E?): E? = this ?: other

如果condition为false则为" error "，否则为"someString"

在编写三元条件运算符之前，让我们考虑以下原型:

if (!answer.isSuccessful()) {
    result = "wrong"
} else {
    result = answer.body().string()
}

return result

解决方案

你可以用!(逻辑上不是)Kotlin if-expression中的运算符:

return if (!answer.isSuccessful()) "wrong" else answer.body().string()

如果你使用if-expression (expression without !操作符):

return if (answer.isSuccessful()) answer.body().string() else "wrong"

Kotlin 's Elvis operator ?:可以做得更好:

return answer.body()?.string() ?: "wrong"

同样，为相应的Answer类使用扩展函数:

fun Answer.bodyOrNull(): Body? = if (isSuccessful()) body() else null

在扩展函数中，由于Elvis操作符，您可以减少代码:

return answer.bodyOrNull()?.string() ?: "wrong"

或者直接用when条件表达式:

when (!answer.isSuccessful()) {
    parseInt(str) -> result = "wrong"
    else -> result = answer.body().string()
}

when替换c类语言的切换操作符。最简单的形式是这样的

when (x) {
    1 -> print("x == 1")
    2 -> print("x == 2")
    else -> {
        print("x is neither 1 nor 2")
    }
}

Java

int temp = a ? b : c;

相当于Kotlin:

var temp = if (a) b else c