NodeJS native module - how to remove debug symbols

Question

I'm writing a native module for Node but would like to remove the debugging information. I'm using node-gyp to build the module.

It's doing a release build, but still, the symbol table is included in the output file.

So I need to remove it with the strip Unix command after the build. Is there a way to remove it in the build itself - ie. specify something in the .gyp file?

Furthermore, even after stripping the debug symbols, I can still use

strings [node-module]

And it lists the names of my functions. Is it possible to remove these also?

This is the command I use to build the native module:

node-gyp rebuild --target=v8.9.4

And this is my binding.gyp:

{
  "targets": [
    {
      "libraries": [
          "/usr/lib/x86_64-linux-gnu/libudev.so",
          "/usr/lib/x86_64-linux-gnu/libboost_regex.so.1.58.0"
      ],
      "target_name": "utils",
      "sources": [ "src/native/utils.cpp" ]
    }
  ]
}

Thanks!

What compiler are You using? Maybe You are not specifying that You don't want **any** debug symbols — bartop, Feb 27 '18 at 08:50
@LachoTomov can you try with the flags `--release --no-debug` — Jiby Jose, Feb 27 '18 at 12:07

Mike Kinghan · Accepted Answer · 2018-02-27T19:15:16.453

As you probably know, a node native module is a dynamic library. You cannot strip the dynamic symbol table from a dynamic library because the dynamic symbol table is necessary at runtime for dynamically linking the library. Even strip --strip-all will not remove the dynamic symbol table. strip --strip-all or strip --strip-unneeded strips everything that can be stripped from a dynamic library. Adding -g0 to the compilation flags eliminates all debugging information, but symbol tables contain more than debugging information.

You can instruct the linker to do the same thing as strip --strip-all at linktime by passing it the option -s|--strip-all. To do this, your binding.gyp would be:

{
    "targets": [
    {
        "libraries": [
            "/usr/lib/x86_64-linux-gnu/libudev.so",
            "/usr/lib/x86_64-linux-gnu/libboost_regex.so.1.58.0"
        ],
        "target_name": "utils",
        "ldflags" : [ "-Wl,-s" ],
        "sources": [ "src/native/utils.cpp" ],
    }
  ]
}

The resulting node module is then as stripped as a shared library can be:

$ file ./build/Release/utils.node
./build/Release/utils.node: ELF 64-bit LSB shared object, x86-64, \
version 1 (SYSV), dynamically linked, \
BuildID[sha1]=eb53cee5839c71b41176bc7a852802035009e8ae, stripped
                                                        ^^^^^^^^

Thanks for the detailed explanation! – LachoTomov Mar 04 '18 at 13:10 — LachoTomov, Mar 04 '18 at 13:10

score 1 · Answer 2 · answered Feb 27 '18 at 13:13

I have searched a bit and there may be a brute force solution of specifying the compiler flag -g0 manually. Take a look here:https://github.com/nodejs/node-gyp/issues/26 . So following Your example and advices given in the github issue the input file should be like this:

{
  "targets": [
    {
      "libraries": [
          "/usr/lib/x86_64-linux-gnu/libudev.so",
          "/usr/lib/x86_64-linux-gnu/libboost_regex.so.1.58.0"
      ],
      "target_name": "utils",
      "cflags_cc": [ "-g0" ]
      "sources": [ "src/native/utils.cpp" ]
    }
  ]
}

Sadly, there is no way for me to test it and it is not platform independant.

Thanks, tried it, plus the different "cflags" variations from the link, but unfortunately still the same. — LachoTomov, Feb 27 '18 at 13:35

NodeJS native module - how to remove debug symbols

2 Answers2