(def x 1)
user=> '`~x
x
user=> `'~x
(quote 1)
Can anyone explain please how it is evaluated step by step?
1 Answers
17
votes
The single-quote operator returns the expression or symbol that you are quoting without evaluating it. The syntax-quote operator returns the expression or symbol that you are quoting (with namespaces added), without evaluating it. The syntax-unquote operator "cancels out" the syntax-quote operator, so to speak, but not the single-quote. You can nest syntax-quotes and syntax-unquotes to perform weird and wonderful feats. My favorite analogy I read for understanding these is to consider syntax-quoting and syntax-unquoting as moving up and down rungs of a ladder (possible source).
In the form `x, x is syntax-quoted, so it isn't evaluated; instead, you get a namespaced symbol (like user/x). But in the form `~x, x is syntax-unquoted again, so it is evaluated:
user=> `~x
1
On to your examples:
Example 1
' is just sugar for (quote ...).
So '`~x becomes (quote `~x). This in turn becomes (quote x) (remember that `~ doesn't really do anything), which is why the whole expression evaluates to the symbol x.
Example 2
In `'~x, let's first replace the ' with quote: `(quote ~x). The expression is syntax-quoted, so it won't be evaluated.
So you can think of the expression (quote ~x) as an "intermediate step." But we're not done. x inside the syntax-quote is syntax-unquoted, just as in my example up above. So even though this expression as a whole won't be evaluated, x will be, and its value is 1. In the end, you get the expression: (quote 1).
Blog post on the topic.
1. – Chiron