狡猾或计划中的define *是什么? [英] What is define* in guile or scheme?
问题描述
我无法通过搜索找到它，guile中的define*是什么?例如，您可以在以下答案中找到它 https://stackoverflow.com/a/24101699/387194
I can't find it by searching, what is define* in guile? You can find it for instance in this answer https://stackoverflow.com/a/24101699/387194
推荐答案
您将在以下文档中找到它:创建高级参数处理过程.
You will find it in the documentation here: Creating advanced argument handling procedures.
6.10.4.1 lambda *并定义*.
lambda *类似于lambda，但有一些扩展名允许可选 和关键字参数.
lambda* is like lambda, except with some extensions to allow optional and keyword arguments.
library syntax: lambda* ([var…]
[#:optional vardef…]
[#:key vardef… [#:allow-other-keys]]
[#:rest var | . var])
body1 body2 …
可选参数和关键字参数也可以具有默认值 当通话中不存在时，通过给出变量的两个元素的列表 名称和表达.例如在
Optional and keyword arguments can also have default values to take when not present in a call, by giving a two-element list of variable name and expression. For example in
(define* (frob foo #:optional (bar 42) #:key (baz 73))
(list foo bar baz))
foo是固定参数，bar是带有默认值的可选参数 值42，而baz是关键字参数，默认值为73.默认 除非需要它们，否则不评估值表达式，直到 该过程被调用.
foo is a fixed argument, bar is an optional argument with default value 42, and baz is a keyword argument with default value 73. Default value expressions are not evaluated unless they are needed, and until the procedure is called.
如果通话中使用的关键字不是这些关键字，通常会出错 由#:key指定，但在定义中添加#:allow-other-keys (在关键字参数声明之后)将忽略unknown 关键字.
Normally it’s an error if a call has keywords other than those specified by #:key, but adding #:allow-other-keys to the definition (after the keyword argument declarations) will ignore unknown keywords.
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