我需要将一个语句列表拆分成多个部分,如下所示:
import macros
macro test: stmt =
var first = quote do:
var x = 1
var second = quote do:
echo x
result = newStmtList()
first.copyChildrenTo(result)
second.copyChildrenTo(result)
echo result.repr
test
但是编译器告诉我这个:
[..]
var x = 1
echo x
minimalist.nim(14, 0) Info: instantiation from here
minimalist.nim(7, 13) Error: undeclared identifier: 'x'
echo x
鉴于在节点列表中声明了x
,这非常令人困惑。我该如何正常工作? (如果它不明显,我确实需要将AST分成多个部分,原因是其他原因)
答案 0 :(得分:3)
默认情况下,Nim会对引用块内的代码强制执行宏卫生规则。这意味着在一个块中引入的符号名称在另一个块中不可见。您可以通过为共享符号引入变量来解决这个问题:
import macros
macro test: stmt =
var x = genSym()
var first = quote do:
var `x` = 1
var second = quote do:
echo `x`
result = newStmtList()
first.copyChildrenTo(result)
second.copyChildrenTo(result)
test
引言是由应用于非脏模板的getAst
操作提供支持。如果您自己使用较低级别的机制,也可以禁用引用的模板的卫生行为,如下所示:
import macros
macro test: stmt =
template first {.dirty.} =
var x = 1
template second {.dirty.} =
echo x
result = newStmtList()
getAst(first()).copyChildrenTo(result)
getAst(second()).copyChildrenTo(result)
test
这将得到相同的结果。请记住,quote
最终将获得一个标志,用于控制引用代码的脏污。
答案 1 :(得分:2)
如果你比较AST,你可以看到出了什么问题:
dumpTree:
var x = 1
echo x
打印:
StmtList
VarSection
IdentDefs
Ident !"x"
Empty
IntLit 1
Command
Ident !"echo"
Ident !"x"
然后在你的例子中:
macro test: stmt =
var first = quote do:
var x = 1
var second = quote do:
echo x
result = newStmtList()
first.copyChildrenTo(result)
second.copyChildrenTo(result)
echo result.treeRepr
test
打印:
StmtList
VarSection
IdentDefs
Sym "x"
Empty
IntLit 1
Command
Sym "echo"
Ident !"x"
注意区别?一次" x"是一个标识,另一个是Sym。不应该查找符号,也许这应该在编译器中更改。但是现在你可以帮助自己并再次用Idents替换所有Syms以强制进行新的查找:
import macros
proc symsToIdents(n): PNimrodNode {.compiletime.} =
if n.kind == nnkSym:
return newIdentNode($n)
result = n
for i in 0 .. <result.len:
result[i] = symsToIdents(n[i])
macro test: stmt =
var first = quote do:
var x = 1
var second = quote do:
echo x
result = newStmtList()
first.copyChildrenTo(result)
second.copyChildrenTo(result)
result = symsToIdents(result)
test
编辑:现在报告为可能的错误:https://github.com/nim-lang/Nim/issues/1843