如何访问 Julia 中通用函数的抽象语法树?
问问题
2230 次
3 回答
5
回顾一下:看起来 Simon 正在寻找与泛型函数关联的特定方法的 AST。我们可以LambdaStaticData
为特定方法获取一个包含 AST 的对象,如下所示:
julia> f(x,y)=x+y
julia> f0 = methods(f, (Any, Any))[1]
((Any,Any),(),AST(:($(expr(:lambda, {x, y}, {{}, {{x, Any, 0}, {y, Any, 0}}, {}}, quote # none, line 1:
return +(x,y)
end)))),())
julia> f0[3]
AST(:($(expr(:lambda, {x, y}, {{}, {{x, Any, 0}, {y, Any, 0}}, {}}, quote # none, line 1:
return +(x,y)
end))))
julia> typeof(ans)
LambdaStaticData
显然,这个 AST 可以是一个 Expr 对象或一个压缩的 AST 对象,表示为一个字节序列:
julia> typeof(f0[3].ast)
Array{Uint8,1}
from的show()
方法说明了遇到这种情况时如何解压:LambdaStaticData
base/show.jl
julia> ccall(:jl_uncompress_ast, Any, (Any, Any), f0[3], f0[3].ast)
:($(expr(:lambda, {x, y}, {{}, {{x, Any, 0}, {y, Any, 0}}, {}}, quote # none, line 1:
return +(x,y)
end)))
julia> typeof(ans)
Expr
于 2013-02-14T20:15:20.517 回答
3
Julia 有四个函数和四个与这些函数类似的宏,用于检查很多关于泛型函数的方法:
julia> f(x, y) = x + y
f (generic function with 1 method)
julia> methods(f)
# 1 method for generic function "f":
f(x,y) at none:1
降低代码:
julia> code_lowered(f, (Int, Int))
1-element Array{Any,1}:
:($(Expr(:lambda, {:x,:y}, {{},{{:x,:Any,0},{:y,:Any,0}},{}}, :(begin # none, line 1:
return x + y
end))))
julia> @code_lowered f(1, 1) # Both `Int`s
...same output.
键入的代码:
julia> code_typed(f, (Int, Int))
1-element Array{Any,1}:
:($(Expr(:lambda, {:x,:y}, {{},{{:x,Int64,0},{:y,Int64,0}},{}}, :(begin # none, line 1:
return (top(box))(Int64,(top(add_int))(x::Int64,y::Int64))::Int64
end::Int64))))
julia> @code_lowered f(1, 1) # Both `Int`s
...same output.
LLVM 代码:
julia> code_llvm(f, (Int, Int))
define i64 @julia_f_24771(i64, i64) {
top:
%2 = add i64 %1, %0, !dbg !1014
ret i64 %2, !dbg !1014
}
julia> @code_llvm f(1, 1) # Both `Int`s
...same output.
本机代码:
julia> code_native(f, (Int, Int))
.text
Filename: none
Source line: 1
push RBP
mov RBP, RSP
Source line: 1
add RDI, RSI
mov RAX, RDI
pop RBP
ret
julia> @code_llvm f(1, 1) # Both `Int`s
...same output.
输入不稳定警告(v0.4+):
julia> @code_warntype f(1, 1)
Variables:
x::Int64
y::Int64
Body:
begin # In[17], line 1:
return (top(box))(Int64,(top(add_int))(x::Int64,y::Int64))
end::Int64
于 2015-02-02T00:20:36.450 回答
0
由于多次分派,我不确定是否存在与通用函数关联的 AST。如果你正在编写一个函数定义fbody
,你应该能够通过做得到 AST dump(quote(fbody))
。
于 2013-01-02T18:51:45.183 回答