Kotlin 当表达式返回函数类型

发布于12月18日

我想利用kotlin的when表达式和通用方法来简化Android的共享首选项api.

我要做的不是一直调用getString()&amp；getInt()等，而是创建一个扩展函数，该函数将根据函数的返回类型进行切换，并调用适当的方法.如下所示:

  fun <T> SharedPreferences.get(key: String): T? {
        when (T) { //how do I switch on return type and call appropriate function?
            is String -> getString(key, null)
            is Int -> getInt(key, -1)
            is Boolean -> getBoolean(key, false)
            is Float -> getFloat(key, -1f)
            is Long -> getLong(key, -1)
        }
        return null
    }

Of Course, it will not work. But is there any solution to use when expression for return type of a function? All suggestions are welcome.

推荐答案

To achieve exactly what you want, you can use reified type parameters. This will make the compiler inline your function at its call sites with T replaced with the type used at the call site.

The function would look like:

@Suppress("IMPLICIT_CAST_TO_ANY")
inline operator fun <reified T> SharedPreferences.get(key: String): T? =
    when (T::class) {
        String::class -> getString(key, null)
        Int::class -> getInt(key, -1)
        Boolean::class -> getBoolean(key, false)
        Float::class -> getFloat(key, -1f)
        Long::class -> getLong(key, -1)
        else -> null
    } as T?

^{If you make get an operator function, you can also call it using the operator syntax: prefs[name].}

The calls should, of course, provide enough type information for the compiler to infer T:

val i: Int? = prefs["i"] // OK, the type information is taken from the declaration
val j: Int = prefs["i"]!! // OK

val x = prefs["x"] // Error, not enough type information
val y = prefs.get<String>("y") // OK, the type will be `String?`

fun f(z: Int) = z
f(prefs["z"]!!) // OK, the type information is taken from the parameter type