为什么嵌套的准引用不能解开拼接-取消引用? [英] Why does the nested quasiquote not unwrap splice-unquote?

问题描述

为什么第一个宏unquote-splicing不产生(quasiquote (unquote (quasiquote 1 2)))，但是第二个示例替换了该值?

Why does the first macro unquote-splicing not produce (quasiquote (unquote (quasiquote 1 2))), but the second example replaces the value?

(define x (list 1 2))

(t '(quasiquote (unquote (quasiquote (unquote-splicing x))))
   (quasiquote (quasiquote (unquote (quasiquote (unquote-splicing x))))))

(t '(quasiquote (unquote (quasiquote (unquote (1 2)))))
   (quasiquote (quasiquote (unquote (quasiquote (unquote (unquote x)))))))

推荐答案

在第一个示例中

(quasiquote (quasiquote (unquote (quasiquote (unquote-splicing x)))))

这有3个准引号和2个非引号，因此不会评估'x'.

This has 3 quasiquotes and 2 unquotes, so 'x' will not be evaluated.

在第二个示例中

(quasiquote (quasiquote (unquote (quasiquote (unquote (unquote x))))))

有3个准引号和3个非引号，因此'x'将被其值替换.

there are 3 quasiquotes and 3 unquotes so 'x' will be replaced by it's value.

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