首先,我不熟悉Linux和进程,所以我不能准确地理解fork()的逻辑。我想从用户输入创建一个进程树,并使用'pstree‘显示此树。但是,我的代码不止一次地显示树。我想原因是fork()复制了'pstree‘命令,我不能解决这个问题。代码是这样的:
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<string>
#include<sys/types.h>
#include<sys/wait.h>
using namespace std;
int main(int argc, char* argv[])
{
int forkSize, currentPid,b=0;
forkSize=atoi(argv[1]);
int parentPid=getpid();
for(int i=0;i<forkSize;i++)
{
currentPid = fork();
if (currentPid<0)
{
cout<<"Error in fork"<<endl;
exit(0);
}
if(currentPid == 0)
{
cout << "Child process: My parent id = " << getppid() << endl;
cout << "Child process: My process id = " << getpid() << endl;
}
else
{
cout << "Parent process. My process id = " << getpid() << endl;
cout << "Parent process. Value returned by fork() = " << currentPid << endl;
}
}
fflush(stdout);
char mychar[500];
sprintf(mychar, "pstree -p -U %d", parentPid);
system(mychar);
fflush(stdout);
cout<<endl;
return 0;
}当我使用输入4尝试此代码时,输出如下所示:
我不明白为什么它会一遍又一遍地显示所有内容。有没有办法只显示一次树?如果能帮到我,我会很高兴的。
发布于 2021-04-07 07:01:19
所有进程最终都会得到运行pstree的代码。您应该在那里检查您是否在原始的父进程中,并且只在这种情况下运行pstree。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<string>
#include<sys/types.h>
#include<sys/wait.h>
using namespace std;
int main(int argc, char* argv[])
{
int forkSize, currentPid,b=0;
forkSize=atoi(argv[1]);
int parentPid=getpid();
for(int i=0;i<forkSize;i++)
{
currentPid = fork();
if (currentPid<0)
{
cout<<"Error in fork"<<endl;
exit(0);
}
if(currentPid == 0)
{
cout << "Child process: My parent id = " << getppid() << endl;
cout << "Child process: My process id = " << getpid() << endl;
sleep(1);
}
else
{
cout << "Parent process. My process id = " << getpid() << endl;
cout << "Parent process. Value returned by fork() = " << currentPid << endl;
}
}
if (getpid() == parentPid) {
char mychar[500];
sprintf(mychar, "pstree -p -U %d", parentPid);
system(mychar);
}
return 0;
}顺便说一句,endl刷新输出,你不需要调用fflush(stdout);。而且,这是C语言,但你的代码是C++。
https://stackoverflow.com/questions/66977343
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