是否有一种轻量级的方法来获取Linux中当前的进程数？

Question

I want my (C/C++ based) program to display a numeric indicator of how many processes are currently present on the local system. The number-of-running-processes value would be queried often (eg once per second) to update my display.

Is there a lightweight way to get that number? Obviously I could call "ps ax | wc -l", but I'd prefer not to force the computer to spawn a process and parse several hundred lines of text just to come up with a single integer.

This program will be running primarily under Linux, but it might also run under MacOS/X or Windows also, so techniques relevant to those OS's would be helpful also.

Ideally I'm looking for something like this , except available under Linux (getsysinfo() appears to be more of a Minix thing).

Answer 1

.... and of course 1 minute after I post the question, I figure out the answer: sysinfo will return (amongst other things) a field that indicates how many processes there are.

That said, if anyone knows of a MacOS/X and/or Windows equivalent to sysinfo(), I'm still interested in that.

---

Update: Here's the function I finally ended up with.

#ifdef __linux__
# include <sys/sysinfo.h>
#elif defined(__APPLE__)
# include <sys/sysctl.h>
#elif defined(WIN32)
# include <Psapi.h>
#endif

int GetTotalNumProcesses()
{
#if defined(__linux__)
   struct sysinfo si;
   return (sysinfo(&si) == 0) ? (int)si.procs : (int)-1;
#elif defined(__APPLE__)
   size_t length = 0;
   static const int names[] = {CTL_KERN, KERN_PROC, KERN_PROC_ALL, 0};
   return (sysctl((int *)names, ((sizeof(names)/sizeof(names[0]))-1, NULL, &length, NULL, 0) == 0) ? (int)(length/sizeof(kinfo_proc)) : (int)-1;
#elif defined(WIN32)
   DWORD aProcesses[1024], cbNeeded;
   return EnumProcesses(aProcesses, sizeof(aProcesses), &cbNeeded) ? (cbNeeded/sizeof(DWORD)) : -1;
#else
   return -1;
#endif
}

Answer 2

opendir("/proc")并计算目录条目数并具有仅数字名称。