execvp返回后，为什么我的程序没有在中断处继续执行? [英] After execvp returns, why doesn't my program pick up where it left off?

问题描述

我有一段像这样的代码作为子线程运行:

I have a block of code like this that runs as a child thread:

if(someVar == 1){
doSomeStuff;

_exit(0)
}
else
   execvp(*(temp->_arguments), temp->_arguments);
printf("I'm done\n");

当我使用someVar == 1运行程序时，我知道_exit(0)调用会杀死我的线程.但是，将其设置为0时，为什么在execvp()调用之后程序不继续运行，而执行printf语句呢?

When I run the program with someVar == 1, I understand that the _exit(0) call kills my thread. However, when it's set to 0, why doesn't the program continue after the execvp() call and do the printf statement?

推荐答案

如果您 exec* (调用exec系列中的任何exec函数)，然后将新程序的代码加载到当前进程中，并继续执行其主要功能及其内容.成功执行这些功能后，它们将永远不会返回，因为您的printf不再存在于内存中.

If you exec* (call any exec function from the exec family), then the code of a new program is loaded into your current process and execution continues with its main function and its stuff. On a successful execution of those functions, they will never return because your printf does not exist anymore in memory.

我认为您将exec*与 fork 混淆了功能.这将拼接出一个新的子进程，该子进程将运行与父进程相同的代码.

I think you confuse exec* with the fork function. That will splice off a new child process which will run the same code as the parent.

如果要创建一个与主线程共享数据和地址空间的新线程，则应使用

If what you want is to create a new thread, that shares data and the address space with the main thread, you should use the pthread_create function. A new process will not share data and you will have to communicate with the other process using other mechanisms, like pipes or shared memory.

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