符号集合转义问题：如何获取变量集合而非转义表达式？

Hey there! The issue with your macro is that the way you're constructing the expression is adding an extra layer of tuple nesting. Let's break this down and fix it.

The Problem #

Your current macro:

macro escape_all(x...) :($(esc.(x))...) end

When expanded, it produces an expression like ((esc(x), esc(y), esc(z))...,), which evaluates to a nested tuple ((1,2,3),) instead of the flat tuple (1,2,3) you're expecting. This happens because the ... in :($(esc.(x))...) is trying to splice the tuple of escaped expressions into an empty context, which wraps it in an extra tuple.

The Solution #

Adjust the macro to explicitly wrap the spliced expressions in a tuple inside the quote:

macro escape_all(x...) :(($(esc.(x))...)) end

Why This Works #

• esc.(x) creates a tuple of escaped expressions for each input variable (to protect them from macro hygiene issues).
• By wrapping $(esc.(x))... in parentheses inside the quote (:(($(esc.(x))...))), we're telling Julia to create a flat tuple expression with each escaped variable as a direct element.
• When macroexpanded, this gives you (esc(x), esc(y), esc(z)), which evaluates directly to the flat tuple of your variables' values.

Testing It Out #

Let's verify with your example:

x = 1
y = 2
z = 3

# Macro expansion now returns a flat tuple of escaped variables
macroexpand(:(@escape_all x y z))  # Output: :(esc(x), esc(y), esc(z))

# Evaluating the macro gives the expected flat tuple
@escape_all x y z  # Output: (1, 2, 3)

That's exactly what you were looking for!

内容的提问来源于stack exchange，提问作者Chris Rackauckas