闭包作为 Kotlin 中默认参数后面的参数

Question

我有一个功能

fun <T> get(path: String, params: MutableMap<String, Any>? = null, headers: MutableMap<String, String>? = null, resolver: ResponseResolver<T>): HttpRequest<T>

哪个 ResponseResolver 是类型别名

typealias ResponseResolver<T> = (HttpResponse) -> T

当我像下面这样调用 get 方法时：

get("/somePath", mutableMapOf("key" to "value")){ httpResponse -> ......some code(Last line is a List<SomeClass>)

然后 Intellij 告诉我

Type inference failed: 

fun <T> get
(
path: String,
params: MutableMap<String, Any>? = ...,
headers: MutableMap<String, String>? = ...,
resolver: ResponseResolver<T> /* = (HttpResponse) → T */
)
: HttpRequest<T>

cannot be applied to
(
String,
MutableMap<String, Any>,
(HttpResponse) → List<SomeClass>
)

我不确定将闭包用作某些具有默认参数的函数的参数是否有任何限制。

Answer 1

Kotlin 不知道 mutableMapOf("key" to "value") 是什么。

弄清楚是params还是headers

get("/somePath", headers = mutableMapOf("key" to "value")){ httpResponse -> ......some code(Last line is a List<SomeClass>)

或

get("/somePath", mutableMapOf<String, Any>("key" to "value")){ httpResponse -> ......some code(Last line is a List<SomeClass>)

闭包作为 Kotlin 中默认参数后面的参数

The closure as parameter behind the default parameter in Kotlin

parameters

lambda

closures

type-inference

kotlin