如何从App1的app2的main（）中触发gdb？

问题描述

提供的app2由app1触发，其中有类似system（./ app2）的内容。 app1也做了很多准备工作，例如创建dirs，文件，配置等等，以便运行app2。

我怎样才能从其主要的第一行gdb app2 ）？

我试过以下的方法无效。

  gdb app2 
b main 
 shell ./app1 
  

解决方案 div>

有几种方法可以实现这一点：

1. 如果 app1 不会关闭 stdin，stdout，stderr ，您可以修改 app1 来代替：`system （gdb ./app2）


2. 您可以修改 app2 为 wait ，例如显示。然后运行 app1 ，并从另一个窗口使用 gdb -p $ child_pid 。
   
3. 如果 app1 没有在 app2 之前分支任何子元素，则可以使用（ gdb）设置follow-fork子并让GDB在 app1 分叉后自动开始调试 app2 它。

Provided app2 is triggered by app1 in which there is something like system("./app2"). app1 also does a lot of preparation such as creating dirs, files, configuration... for app2 running.

How can I gdb app2 from the first line of its main()?

What I tried as below doesn't work.

gdb app2
  b main
  shell ./app1

解决方案

There are a few ways to achieve this:

1. If app1 doesn't close stdin, stdout, stderr, you could modify app1 to do this instead: `system("gdb ./app2")
2. You could modify app2 to wait for debugger to be attached, as e.g. this answer shows. Then run app1, and use gdb -p $child_pid from another window.
3. If app1 doesn't fork any children prior to app2, you could use (gdb) set follow-fork child and have GDB automatically start debugging app2 after app1 forks it.

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