通过外部函数或Julia中的宏获取执行文件的路径 [英] get path of the executed file through external function or macro in Julia

问题描述

我正在尝试编写一个没有参数的辅助函数或宏，该函数可以记录文件名和调用它的行.

I'm trying write a helper function or macro with no arguments that can record the filename and line of where it is called.

该帮助器位于另一个模块中，并导入到脚本中，因此@__FILE__和@__LINE__不会指向正确的位置.

The helper is located in a different module and imported to a script, so @__FILE__ and @__LINE__ would not point to the right place.

这是我在trace.jl的帮助程序模块:

Here is my helper module at trace.jl:

module Trace
export @trace, Location

struct Location
    file:: String
    line:: Integer
end

macro trace()
    return Location(abspath(PROGRAM_FILE), __source__.line)
end    

end

这是脚本caller.jl

include("trace.jl")
using .Trace

# putting two statements in one line so that line number is the same
println("I want: ", Location(@__FILE__, @__LINE__)); println(" I get: ", @trace)

运行julia caller.jl的输出如下:

D:\github\Handout.jl\src>julia caller.jl
I want: Location("D:\\github\\Handout.jl\\src\\caller.jl", 5)
 I get: Location("D:\\github\\Handout.jl\\src\\caller.jl", 5)

我不确定PROGRAM_FILE是不是偶然提供了我caller.jl还是可以提供更多保证?

I'm not sure if PROGRAM_FILE provides me caller.jl by accident or there can be more guarantee?

我本来会更乐于从__source__.file提取路径，因为__source__.line将我指向确切文件中的确切行，但是当我尝试时__source__.file是nothing.

I would have been happier to extract path from __source__.file because __source__.line points me to the exact line in exact file, but __source__.file is nothing when I tried.

这里是文档中的两部分. 第一个:

Here are two pieces in the documentation. The first one:

除了给定的参数列表外，每个宏还传递了名为__source__和__module__的其他参数.

自变量__source__通过宏调用提供有关@ sign的解析器位置的信息(以LineNumberNode对象的形式).

The argument __source__ provides information (in the form of a LineNumberNode object) about the parser location of the @ sign from the macro invocation.

第二个:

源位置信息表示为(line line_num file_name)，其中第三部分是可选的(当当前行号(但文件名未更改)时省略).

Source location information is represented as (line line_num file_name) where the third component is optional (and omitted when the current line number, but not file name, changes).

这些表达式在Julia中表示为LineNumberNode.

These expressions are represented as LineNumberNodes in Julia.

是否有可能攀登LineNumberNode链以获得文件名而不是nothing?

Is there possibly a way to climb up LineNumberNode chain to get a filename and not nothing?

也许还有一种方法可以将%__FILE__的计算延迟到运行时，以便我可以在trace中使用该构造?

Also maybe there is a way to delay computation of %__FILE__ until runtime, so that I can use that construct in trace?

类似的讨论: Julia:创建一个相对于脚本位置的新文件夹和文件

推荐答案

引用__source__是Julia手册中建议的内容.这是一个例子

Quoting __source__ is what is recommended in the Julia manual. Here is an example

module Trace

export @trace

macro trace()
    return QuoteNode(__source__)
end

end # module

文件f2.jl

include("f1.jl")

using .Trace

println("I want: ", (@__FILE__, @__LINE__)); println("I get: ", @trace)

x = @trace
dump(x)

println("This is not what you want: ", PROGRAM_FILE)

文件f3.jl

include("f2.jl")

运行以上

现在看一下输出:

Running the above

Now have a look at the output:

$ julia f3.jl
I want: ("D:\\f2.jl", 5)
I get: #= D:\f2.jl:5 =#
LineNumberNode
  line: Int64 7
  file: Symbol D:\f2.jl
This is not what you want: f3.jl

尤其是:

• @trace返回具有两个字段的LineNumberNode对象(但我知道这是您想要的)
• 您会看到PROGRAM_FILE为您提供了不同的信息:这是从命令行传递给Julia的文件的名称(因此在我们的情况下为f3.jl，尽管在​​f2.jl文件中被调用了是f3.jl的include d.)

• @trace returns you LineNumberNode object that has two fields (but I understand this is what you want)
• you can see that PROGRAM_FILE gives you a different information: it is a name of the file passed to Julia from the command line (so it is f3.jl in our case, although it was called in f2.jl file which was included by f3.jl).

现在更清楚吗?

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