为什么返回类型是函数＆lt; t，r＆gt;在Groovy中忽略？

发布于 2025-01-29 11:54:24 字数 387 浏览 2 评论 0原文

以下简单代码解释了我的困惑：

class Main {

    static void f(Function<Float, Float> c) {
        println(c.apply(0.0f))
    }

    static void main(String[] args) {
        Closure<String> c = {"hi"}
        f(c)
    }

}

我不知道为什么编译器不抱怨cloture＆lt; string＆gt;不适合function＆lt; float，float＆gt;。似乎我可以将任何内容传递给f（）。

原文

The following simple code explains my confusion:

class Main {

    static void f(Function<Float, Float> c) {
        println(c.apply(0.0f))
    }

    static void main(String[] args) {
        Closure<String> c = {"hi"}
        f(c)
    }

}

I have no idea why the compiler does not complain that Closure<String> is not appropriate for Function<Float, Float>. Seems that I can pass anything to f().

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柠檬色的秋千 2025-02-05 11:54:24

以下代码

import java.util.function.*

def c = {"result $it :: ${it.getClass()}"}
Function<Float, Float> f = c

println "f: ${f.getClass()}   ${f instanceof Function}"
println "c: ${c.getClass()}   ${c instanceof Function}"

println f.apply(0.1)

打印

f: class com.sun.proxy.$Proxy22   true
c: class ConsoleScript10$_run_closure1   false
result 0.1 :: class java.math.BigDecimal

否则没有类型检查
groovy是动态的 - 除非您指定此（编译）闭合没有实现功能，。因此，当您将闭合分配到函数时 - Groovy试图通过功能接口委托闭合。即使您使用编译静态...

the following code

import java.util.function.*

def c = {"result $it :: ${it.getClass()}"}
Function<Float, Float> f = c

println "f: ${f.getClass()}   ${f instanceof Function}"
println "c: ${c.getClass()}   ${c instanceof Function}"

println f.apply(0.1)

prints

f: class com.sun.proxy.$Proxy22   true
c: class ConsoleScript10$_run_closure1   false
result 0.1 :: class java.math.BigDecimal

groovy is dynamic - there is no type check unless you specify this (CompileStatic)
closure does not implement function. so, when you are assigning closure into function - groovy tries to delegate closure through a function interface. it will be dynamic even if you use compile static...

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