scheme - 语法规则表达式本身是否在 Scheme 中计算为一个值？

Question

在 Chibi 和 CHICKEN 中，以下 syntax-rules 表达式计算为一个过程:

(syntax-rules () ((_) #f))

这只是这些特定实现的编写方式的产物吗？ Scheme 语言规范似乎没有指出 syntax-rules 能够计算出一个值。

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更新

这似乎取决于Scheme的版本？

From the R⁶RS Spec :

Semantics: An instance of syntax-rules evaluates, at macro-expansion time, to a new macro transformer by specifying a sequence of hygienic rewrite rules. A use of a macro whose keyword is associated with a transformer specified by syntax-rules is matched against the patterns contained in the s, beginning with the leftmost . When a match is found, the macro use is transcribed hygienically according to the template. It is a syntax violation when no match is found.

来自R⁵RS Spec和 R⁷RS Spec :

Semantics: An instance of syntax-rules produces a new macro transformer by specifying a sequence of hygienic rewrite rules. A use of a macro whose keyword is associated with a transformer specified by syntax-rules is matched against the patterns contained in the syntax rules, beginning with the leftmost syntax rule. When a match is found, the macro use is transcribed hygienically according to the template.

Answer 1

syntax-rules返回一个 transformer，它是 expander 用来将句法扩展转换为另一个句法扩展的过程。另见 here .

所以不，它不是一种特殊形式(同时它已经从你的问题中消失了)，是的，它的计算结果是一个值，因为 Scheme 有 first-class procedures因此程序就是值(value)。

这是 Scheme 的标准行为，而不是 Chicken 或 Chibi 的实现细节。