如何在不调试符号的情况下获取OCaml中的异常的行号? [英] How to get the line number of an exception in OCaml without debugging symbols?

问题描述

有没有一种无需调试符号即可获取OCaml中异常行号的好方法吗?当然，如果打开调试符号并使用OCAMLRUNPARAM=b运行，则可以获取回溯信息.但是，我实际上并不需要整个回溯，因此我想一个无需调试符号的解决方案.目前，我们可以编写类似

Is there a good way to get the line number of exception in OCaml without debugging symbols? Certainly, if we turn on debugging symbols and run with OCAMLRUNPARAM=b, we can get the backtrace. However, I don't really need the whole backtrace and I'd like a solution without debugging symbols. At the moment, we can write code like

try
    assert false
with x ->
    failwith (Printexc.to_string x ^ "\nMore useful message")

以便从assert中获取文件和行号，但这似乎很尴尬.有没有更好的方法来获取异常的文件和行号?

in order to get the file and line number from assert, but this seems awkward. Is there a better way to get the file and line number of the exception?

推荐答案

在任何地方都可以使用全局符号__FILE__和__LINE__.

There are global symbols __FILE__ and __LINE__ that you can use anywhere.

$ ocaml
        OCaml version 4.02.1

# __FILE__;;
- : string = "//toplevel//"
# __LINE__;;
- : int = 2
# 

更新

正如@MartinJambon所指出的，还有__LOC__，它在一个字符串中给出了文件名，行号和字符位置:

As @MartinJambon points out, there is also __LOC__, which gives the filename, line number, and character location in one string:

# __LOC__;;
- : string = "File \"//toplevel//\", line 2, characters -9--2"

更新2

这些符号在 Pervasives模块中定义.完整列表为:__LOC__，__FILE__，__LINE__，__MODULE__，__POS__，__LOC_OF__，__LINE_OF__，__POS_OF__.

These symbols are defined in the Pervasives module. The full list is: __LOC__, __FILE__, __LINE__, __MODULE__, __POS__, __LOC_OF__, __LINE_OF__, __POS_OF__.

最后三个返回有关整个表达式而不是文件中单个位置的信息:

The last three return information about a whole expression rather than just a single location in a file:

# __LOC_OF__ (8 * 4);;
- : string * int = ("File \"//toplevel//\", line 2, characters 2-9", 32)

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