通过多个过程从stdin读取 [英] Reading from stdin by multiple processes
问题描述
我正在编写一个程序,其中父进程使用fork()
创建N
子进程(将N作为参数提供),以便每个子进程都直接由这个父进程派生.
I am writing a program in which the parent process uses fork()
to create N
child processes (N is provided as an argument), so that every child is directly forked by this one parent.
每个子进程都需要从stdin
中读取一行,并打印到屏幕上.
编译并执行程序后,将通过以下文本文件提供文本:
Every child process needs to read a line from stdin
, and print on to the screen.
After compiling and executing the program, the text is provided through a text file like this:
./prog1 3 < fileWithText.txt
我的问题是,我希望看到每个孩子都在为阅读输入而打架",但是我实际上看到的是,总是只有一个子进程负责处理输入.这是我用于子进程的代码:
My problem is, that I am expecting to see every child "fight" for reading the input, however what I actually see is that there is always only one child process taking care of handling the input. This is the code I use for the child process:
void do_child_reader(int j) {
int pid;
int counter = 0;
char* buffer;
char* readCheck;
int readerRunning = TRUE;
pid = getpid();
buffer = (char*)malloc(BUFFER_SIZE * sizeof(char));
while (readerRunning == TRUE) {
readCheck = fgets(buffer, BUFFER_SIZE, stdin);
if (readCheck == NULL) {
readerRunning = FALSE;
}
else {
fprintf(stdout, "(READER %d pid-%d) %s", j, pid, buffer);
counter++;
}
}
fprintf(stderr, "(READER %d pid-%d) processed %d messages, going to exit\n", j, pid, counter);
free(buffer);
exit(counter);
}
这是我在运行N = 3时得到的输出:
Here's the output I get when running for N=3:
(READER 0 pid-72655) I am a message - line 1
(READER 0 pid-72655) I am a message - line 2
(READER 0 pid-72655) I am a message - line 3
(READER 0 pid-72655) processed 3 messages, going to exit
(READER 1 pid-72657) processed 0 messages, going to exit
(READER 2 pid-72659) processed 0 messages, going to exit
父进程正在等待子进程完成后退出.
The parent process is waiting for the children to finish before exiting.
我正在尝试找出导致此行为的原因,以及它是通过fgets()
的方式还是以我尝试读取字符串的方式.
I am trying to figure out what would cause this behaviour, and whether it is in the way fgets()
works or perhaps in the way I am trying to read the strings.
谢谢!
推荐答案
如果文件太小,一个进程可以读取一个BUFSIZ
字节(来自<stdio.h>
)字节,那么其他进程就什么也没剩下读书.增大输入文件的大小(是BUFSIZ
大小的倍数),并确保客户机在缓冲区已满后读取速度足够慢,并且您会看到它们都在工作.
If the file is so small that one unit of BUFSIZ
(from <stdio.h>
) bytes can be read by one process, the other processes have nothing left to read. Make the input file bigger — multiple times the size of BUFSIZ
— and make sure the clients read slowly enough after getting a buffer full, and you will see them all working on it.
这是为您提供的I/O缓冲.您可以尝试弄乱setvbuf()
来设置输入的行缓冲.我不确定是否可以.
This is I/O buffering for you. You could try messing with setvbuf()
to set line buffering on the input; I'm not sure it would work.
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