没有控制台的C++ popen命令

Question

当我使用popen来获取命令的输出时,比如dir,它会提示出一个控制台.

但是,如果没有控制台的外观,我可以获得命令的输出吗？

我正在使用Visual C++,并希望使一个库返回一些命令的输出,比如dir.

Answer 1

假设Windows(因为这是此行为流行的唯一平台):

CreatePipe()用于创建通信所需的管道,CreateProcess用于创建子进程.

HANDLE StdInHandles[2]; 
HANDLE StdOutHandles[2]; 
HANDLE StdErrHandles[2]; 

CreatePipe(&StdInHandles[0], &StdInHandles[1], NULL, 4096); 
CreatePipe(&StdOutHandles[0], &StdOutHandles[1], NULL, 4096); 
CreatePipe(&StdErrHandles[0], &StdErrHandles[1], NULL, 4096); 


STARTUPINFO si;   memset(&si, 0, sizeof(si));  /* zero out */ 

si.dwFlags =  STARTF_USESTDHANDLES; 
si.hStdInput = StdInHandles[0];  /* read handle */ 
si.hStdOutput = StdOutHandles[1];  /* write handle */
si.hStdError = StdErrHandles[1];  /* write handle */

/* fix other stuff in si */

PROCESS_INFORMATION pi; 
/* fix stuff in pi */


CreateProcess(AppName, commandline, SECURITY_ATTRIBUTES, SECURITY_ATTRIBUTES, FALSE, CREATE_NO_WINDOW |DETACHED_PROCESS, lpEnvironment, lpCurrentDirectory, &si, &pi); 

这不仅仅是让你顺利完成你想要完成的任务.

Answer 2

对于 POSIX，它应该是这样的：

//Create the pipe.
int lsOutPipe[2];
pipe(lsOutPipe);

//Fork to two processes.
pid_t lsPid=fork();

//Check if I'm the child or parent.
if ( 0 == lsPid )
{//I'm the child.
  //Close the read end of the pipe.
  close(lsOutPipe[0]);

  //Make the pipe be my stdout.
  dup2(lsOutPipe[1],STDOUT_FILENO);

  //Replace my self with ls (using one of the exec() functions):
  exec("ls"....);//This never returns.  
} // if

//I'm the parent.
//Close the read side of the pipe.
close(lsOutPipe[1]);

//Read stuff from ls:
char buffer[1024];
int bytesRead;
do
{
  bytesRead = read(emacsInPipe[0], buffer, 1024);

  // Do something with the read information.
  if (bytesRead > 0) printf(buffer, bytesRead);
} while (bytesRead > 0);

您当然应该检查返回值等......