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c++ 程序如何在调试时读取 VSCode 中的用户输入

[英]How c++ program read user input in VSCode while debugging

I'm trying to do debugging c/c++ program in vscode debug tab.我正在尝试在 vscode 调试选项卡中调试 c/c++ 程序。 Well, after setting some properties which is spread through google, it works well.好吧,在设置了一些通过谷歌传播的属性后,它运行良好。 However, when I try to do debug with program containing "scanf" or "cin" which requires user input, it never receives any input from my keyboard.但是,当我尝试使用包含需要用户输入的“scanf”或“cin”的程序进行调试时,它永远不会从我的键盘接收到任何输入。 How can I do so?我该怎么做?

And what I found from google is that setting "externalConsole" value as true like this picture.我从谷歌上发现的是像这张图片一样将“externalConsole”值设置为

{
  "version": "0.2.0",
  "configurations": [
    {
      "name": "g++ - 활성 파일 빌드 및 디버그",
      "type": "cppdbg",
      "request": "launch",
      "program": "${workspaceFolder}/Sort/test",
      "args": [],
      "stopAtEntry": false,
      "cwd": "${workspaceFolder}",
      "environment": [],
      "externalConsole": true,
      "MIMode": "lldb",
      "preLaunchTask": "C/C++: g++ build active file"
    }
  ]
}

However, if I change the value of externalConsole as true and try to debug program containing cin, there is no progress after the below picture.但是,如果我将 externalConsole 的值更改为 true 并尝试调试包含 cin 的程序,则在下图之后没有任何进展。

尝试调试给定程序的那一刻

Even though I try to input the data in the external console(for example "asd"), it says "zsh:command not found asd".即使我尝试在外部控制台中输入数据(例如“asd”),它也会显示“zsh:command not found asd”。 How can I fix it for program to receive my input.我该如何修复它以便程序接收我的输入。 HELP ME PLZ请帮帮我

  1. install CodeRunner extension on your VSCode在你的 VSCode 上安装 CodeRunner 扩展
  2. and then follow steps: Code -> Preferences -> Settings然后按照以下步骤操作: Code -> Preferences -> Settings
  3. then on setting type CodeRunner Settings然后在设置类型 CodeRunner Settings
  4. find the Run In Terminal找到运行终端
  5. Enable that function启用 function

Follow these steps and you will be able to run your code on VSCode terminal and it would be possible to enter Input按照这些步骤,您将能够在 VSCode 终端上运行您的代码,并且可以输入Input

enable this启用这个

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