为什么Scala编译器不允许使用默认参数重载方法? [英] Why does the Scala compiler disallow overloaded methods with default arguments?

问题描述

虽然在某些情况下这种方法的重载可能会变得模棱两可，但是为什么编译器会不允许在编译时和运行时既不模棱两可的代码呢?

While there might be valid cases where such method overloadings could become ambiguous, why does the compiler disallow code which is neither ambiguous at compile time nor at run time?

示例:

// This fails:
def foo(a: String)(b: Int = 42) = a + b
def foo(a: Int)   (b: Int = 42) = a + b

// This fails, too. Even if there is no position in the argument list,
// where the types are the same.
def foo(a: Int)   (b: Int = 42) = a + b
def foo(a: String)(b: String = "Foo") = a + b

// This is OK:
def foo(a: String)(b: Int) = a + b
def foo(a: Int)   (b: Int = 42) = a + b    

// Even this is OK.
def foo(a: Int)(b: Int) = a + b
def foo(a: Int)(b: String = "Foo") = a + b

val bar = foo(42)_ // This complains obviously ...

有什么理由不能放宽这些限制吗?

Are there any reasons why these restrictions can't be loosened a bit?

特别是在将重载的Java代码转换为Scala时，默认参数非常重要，并且在用规范/编译器施加了任意限制的一种Scala方法替换了许多Java方法之后，很难找到该参数.

Especially when converting heavily overloaded Java code to Scala default arguments are a very important and it isn't nice to find out after replacing plenty of Java methods by one Scala methods that the spec/compiler imposes arbitrary restrictions.

推荐答案

我想引用Lukas Rytz(摘自

I'd like to cite Lukas Rytz (from here):

原因是我们想要确定性的命名方案 生成的方法返回默认参数.如果你写

The reason is that we wanted a deterministic naming-scheme for the generated methods which return default arguments. If you write

def f(a: Int = 1)

编译器生成

def f$default$1 = 1

如果在同一参数上有两个具有默认值的重载 位置，我们需要一个不同的命名方案.但是我们要保持 生成的字节码在多个编译器运行中都稳定.

If you have two overloads with defaults on the same parameter position, we would need a different naming scheme. But we want to keep the generated byte-code stable over multiple compiler runs.

未来Scala版本的解决方案可能是将非默认参数的类型名称(在方法的开头，它们使重载的版本消除歧义)合并到命名模式中，例如在这种情况下:

A solution for future Scala version could be to incorporate type names of the non-default arguments (those at the beginning of a method, which disambiguate overloaded versions) into the naming schema, e.g. in this case:

def foo(a: String)(b: Int = 42) = a + b
def foo(a: Int)   (b: Int = 42) = a + b

那会是这样的:

def foo$String$default$2 = 42
def foo$Int$default$2 = 42

有人愿意编写SIP提案?

这篇关于为什么Scala编译器不允许使用默认参数重载方法?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！