当前位置：文江博客话题详情

不同编程语言中的重载

发布于 2024-10-11 06:08:25 字数 39 浏览 4 评论 0原文

有人可以解释（举例）上下文无关重载和上下文相关重载之间的区别吗？

原文

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

谈下烟灰 2024-10-18 06:08:25

我从来没有听说过这些。谷歌上只有大约五次点击，其中之一就是这个问题，这似乎表明这些都是虚构的术语。与任何虚构的术语一样，如果你想知道它的含义，你必须询问创造它的人。

据我所知，它似乎与基于返回类型的重载有关。

基本上，如果您有四个像这样的重载函数：

foo :: string → int
foo :: string → string
foo :: string → ()
foo :: int → int

这样调用它们：

1 + foo 1
1 + foo "one"
foo "one"

然后，使用上下文相关的重载（即基于返回类型和参数类型的重载），将选择以下实现：

1 + foo 1     # foo :: int → int
1 + foo "one" # foo :: string → int (because `+` takes an `int`)
foo "one"     # foo :: string → ()  (because there is no return value)

并且您与上下文无关的重载（即忽略返回类型），将选择以下实现：

1 + foo 1     # foo :: int → int
1 + foo "one" # ERROR
foo "one"     # ERROR

在两种 ERROR 情况下，foo :: string → int 之间存在歧义、 foo :: string → string 和 foo :: string → ()，因为它们仅在返回类型上有所不同，但具有相同的参数类型。

I have never heard about those. And there's only about five hits on Google, one of which is this very question, which seems to suggest to me that these are made-up terms. And as with any made-up term, if you want to know what it means, you have to ask the person who made it up.

From what little I could gather, it seems to be related to return-type based overloading.

Basically, if you have four overloaded functions like these:

foo :: string → int
foo :: string → string
foo :: string → ()
foo :: int → int

And you call them like this:

1 + foo 1
1 + foo "one"
foo "one"

Then, with context-dependent overloading (i.e. overloading based on the return type as well as the parameter types), the following implementations will be selected:

1 + foo 1     # foo :: int → int
1 + foo "one" # foo :: string → int (because `+` takes an `int`)
foo "one"     # foo :: string → ()  (because there is no return value)

Whereas with context-independent overloading (i.e. ignoring the return type), the following implementations will be selected:

1 + foo 1     # foo :: int → int
1 + foo "one" # ERROR
foo "one"     # ERROR

In both the ERROR cases, there is an ambiguity between foo :: string → int, foo :: string → string and foo :: string → (), since they only differ in their return type but have the same paremeter type.

回复收藏 0 原文