如何在VS2022中调试WSL Ubuntu下用make编译的C++项目？

问题背景 #

必须在Ubuntu WSL环境下用make编译C++项目，同时使用VS2022作为IDE，且只能采用无.sln/.vcxproj的文件夹式项目。配置tasks.vs.json和launch.vs.json后，调试时出现错误。

错误信息 #

Starting: "wsl" "/usr/bin/gdb --interpreter=mi"
/bin/bash: line 1: /usr/bin/gdb --interpreter=mi: No such file or directory
"wsl" exited with code 127 (0x7F).

原配置文件 #

launch.vs.json #

{
  "version": "0.2.1",
  "defaults": {},
  "configurations": [
    {
      "type": "cppdbg",
      "name": "WSL Launch",
      "request": "launch",
      "project": "main.exe",
      "program": "/mnt/c/Task1/main.exe",
      "args": [],
      "stopAtEntry": false,
      "cwd": "/mnt/c/Task1",
      "environment": [],
      "externalConsole": false,
      "pipeTransport": {
        "pipeCwd": "",
        "pipeProgram": "wsl",
        "pipeArgs": [],
        "debuggerPath": "/usr/bin/gdb"
      },
      "MIMode": "gdb",
      "setupCommands": [
        {
          "description": "Enable pretty-printing for gdb",
          "text": "-enable-pretty-printing",
          "ignoreFailures": true
        }
      ],
      "preLaunchTask": "build"
    }
  ]
}

tasks.vs.json #

{
  "version": "0.2.1",
  "tasks": [
    {
      "taskLabel": "Task-2024-14-10",
      "appliesTo": "/",
      "type": "launch"
    },
    {
      "taskName": "build",
      "appliesTo": ".*",
      "type": "command",
      "command": "wsl",
      "args": [
        "make"
      ],
      "label": "Build with WSL"
    },
    {
      "taskName": "run",
      "appliesTo": ".*",
      "type": "command",
      "command": "wsl",
      "args": [
        "./main.exe"
      ],
      "label": "Run Application in WSL"
    },
    {
      "taskLabel": "Task-2024-14-10",
      "appliesTo": "/",
      "type": "launch"
    },
    {
      "taskLabel": "Task-main",
      "appliesTo": "main.cpp",
      "type": "launch"
    },
    {
      "taskLabel": "Task-\u00DCbung 1",
      "appliesTo": "/",
      "type": "launch"
    }
  ]
}

问题分析 #

1. gdb调用参数格式错误：launch配置中pipeTransport.pipeArgs为空，导致VS将/usr/bin/gdb --interpreter=mi作为一个完整的命令字符串传给WSL。WSL会尝试查找名为/usr/bin/gdb --interpreter=mi的可执行文件，自然找不到，触发127错误。
2. 可执行文件路径错误：Linux环境下编译的程序默认无.exe后缀，原配置中program和run任务的args使用.exe后缀，会导致找不到编译产物。
3. 无效配置冗余：tasks.vs.json中存在多个重复无用的launch类型任务，会干扰配置解析。

修复方案 #

修正launch.vs.json #

{
  "version": "0.2.1",
  "defaults": {},
  "configurations": [
    {
      "type": "cppdbg",
      "name": "WSL Launch",
      "request": "launch",
      "program": "/mnt/c/Task1/main", // 去掉.exe后缀，匹配Linux编译产物
      "args": [],
      "stopAtEntry": false,
      "cwd": "/mnt/c/Task1",
      "environment": [],
      "externalConsole": false,
      "pipeTransport": {
        "pipeCwd": "",
        "pipeProgram": "wsl",
        "pipeArgs": ["--", "/usr/bin/gdb", "--interpreter=mi"], // 拆分gdb命令与参数
        "debuggerPath": "/usr/bin/gdb"
      },
      "MIMode": "gdb",
      "setupCommands": [
        {
          "description": "Enable pretty-printing for gdb",
          "text": "-enable-pretty-printing",
          "ignoreFailures": true
        }
      ],
      "preLaunchTask": "build"
    }
  ]
}

• 移除无效的"project": "main.exe"字段
• 修改program路径为Linux下的可执行文件（假设make编译出的是main）
• 给pipeArgs添加["--", "/usr/bin/gdb", "--interpreter=mi"]，让WSL正确拆分执行gdb命令

修正tasks.vs.json #

{
  "version": "0.2.1",
  "tasks": [
    {
      "taskName": "build",
      "appliesTo": ".*",
      "type": "command",
      "command": "wsl",
      "args": [
        "make"
      ],
      "label": "Build with WSL"
    },
    {
      "taskName": "run",
      "appliesTo": ".*",
      "type": "command",
      "command": "wsl",
      "args": [
        "./main" // 去掉.exe后缀
      ],
      "label": "Run Application in WSL"
    }
  ]
}

• 删除所有无用的重复launch类型任务
• 修改run任务的args为./main，匹配Linux编译产物

可行性说明 #

这种无.sln/.vcxproj的文件夹式项目完全可行。VS2022支持通过自定义tasks和launch配置，直接对接WSL环境的make编译和gdb调试，无需创建正式VS项目，只要配置符合WSL和cppdbg的要求即可正常工作。

内容的提问来源于stack exchange，提问作者Bugsia