如何插入Julia"for"表达式？

Question

我正在编写一个基于Python的列表推导的宏@vcomp(矢量理解)和一个条件子句,以简洁的方式过滤元素.

macro vcomp(comprehension::Expr, when::Symbol, condition)
    comp_head, comp_args = comprehension.head, comprehension.args
    comp_head ? [:comprehension, :typed_comprehension] && error("@vcomp not a comprehension")
    when ? :when && error("@vcomp expected `when`, got: `$when`")
    T = comp_head == :typed_comprehension ? comp_args[1] : nothing
    if VERSION < v"0.5-"
        element  = comp_head == :comprehension ? comp_args[1] : comp_args[2]
        sequence = comp_head == :comprehension ? comp_args[2] : comp_args[3]
    else
        element  = comp_head == :comprehension ? comp_args[1].args[1] : comp_args[2].args[1]
        sequence = comp_head == :comprehension ? comp_args[1].args[2] : comp_args[2].args[2]
    end
    result = T ? nothing ? :($T[]) : :([])
    block = Expr(:let, Expr(:block,
                Expr(:(=), :res, result),
                Expr(:for, sequence,
                    Expr(:if, condition,
                        Expr(:call, :push!, :res, element))),
                :res))
    return esc(block)
end

像这样使用:

julia> @vcomp Int[i^3 for i in 1:10] when i % 2 == 0
5-element Array{Int64,1}:
    8
   64
  216
  512
 1000

其中扩展到:

julia> macroexpand(:(@vcomp Int[i^3 for i in 1:15] when i % 2 == 0))
:(let
        res = Int[]
        for i = 1:15
            if i % 2 == 0
                push!(res,i ^ 3)
            end
        end
        res
    end)

我原以为能够block像这样写:

block = quote
    let
        res = $result
        for $sequence
            if $condition
                push!(res, $element)
            end
        end
        res
    end
end

这给出了以下错误:

• ERROR: syntax: invalid iteration specification

而不是我提出的方式:

block = Expr(:let, Expr(:block,
            Expr(:(=), :res, result),
            Expr(:for, sequence,
                Expr(:if, condition,
                    Expr(:call, :push!, :res, element))),
            :res))

但是我能够Expr(:for, ...)直接使用如上所示,据我所知,这是一个解析器错误(这是一个错误？).我也无法找到这种插值的例子,这是我尝试过的:

julia> ex? = :(i in 1:10)
:($(Expr(:in, :i, :(1:10))))

julia> ex? = :(i = 1:10)
:(i = 1:10)

julia> quote
           for $ex?
ERROR: syntax: invalid iteration specification

julia> quote
           for $ex?
ERROR: syntax: invalid iteration specification

构造整个表达式并检查它:

julia> ex? = quote
           for i in 1:10
               print(i)
           end
       end
quote  # none, line 2:
    for i = 1:10 # none, line 3:
        print(i)
    end
end

julia> ex?.args
2-element Array{Any,1}:
 :( # none, line 2:)
 :(for i = 1:10 # none, line 3:
        print(i)
    end)

julia> ex?.args[2].args
2-element Array{Any,1}:
 :(i = 1:10)
 quote  # none, line 3:
    print(i)
end

julia> ex?.args[2].args[1]
:(i = 1:10)

julia> ex?.args[2].args[1] == ex?    # what's the difference then?
true

这有效,但不太可读:

julia> ex? = Expr(:for, ex?, :(print(i)))
:(for $(Expr(:in, :i, :(1:10)))
        print(i)
    end)

julia> ex? = Expr(:for, ex?, :(print(i)))
:(for i = 1:10
        print(i)
    end)

julia> eval(ex?)
12345678910
julia> eval(ex?)
12345678910    
julia> eval(ex?)
12345678910    

有没有办法可以使用更简洁的语法？我发现当前的实现难以阅读和理由与我期望写的相比.

Answer 1

首先,我相信对守卫的理解是来到朱莉娅(在v0.5？).

回答你的问题:解析器希望能够验证其输入在语法上是否正确,而无需查看插值的实际值.试试例如

x, y = :i, :(1:10)
quote
    for $x = $y
    end
end

现在,解析器可以识别语法的相关部分.(如果你使用的话,你应该得到相同的AST for $x in $y.)