什么样的方法是正确的解决这个问题?
例如,我有一个名为write.c
的程序,它有 4 个子进程,子进程将其 PID 写入单个全局命名管道。
另一个名为read.c
的程序应读取此 PID。
我有一个如下所示的方法,但这种方法存在一些问题。它不能读取所有PID,有时是其中的3个,有时是其中的2个。我认为存在同步问题,如何解决这个问题?:
writer.c
:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <fcntl.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
int main(){
int fd;
char * myfifo = "/tmp/myfifo"; //FIFO file
char buffer[50];
mkfifo(myfifo, 0666); //creating the FIFO
for(int i=0;i<4;i++){ //creating 4 child process
if(fork() == 0) {
fd = open(myfifo, O_WRONLY); //each child process opens the FIFO for writing their own PID.
sprintf(buffer, "%d", getpid()); //each child process gets pid and assign it to buffer
printf("write:%sn", buffer); // each child process prints to see the buffer clearly
write(fd, buffer, strlen(buffer)+1); //each child process writes the buffer to the FIFO
close(fd);
exit(0);
}
}
for(int i=0;i<4;i++) { //waiting the termination of all 4 child processes.
wait(NULL);
}
//parent area
}
reader.c
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/stat.h>
#include <time.h>
#include <string.h>
#include <fcntl.h>
int main(int argc, char **argv) {
int fd1;
// FIFO file path
char * myfifo = "/tmp/myfifo";
// Creating the named file(FIFO)
mkfifo(myfifo, 0666);
char str1[80]; //str2[80];
while (1)
{
// First open in read only and read
fd1 = open(myfifo,O_RDONLY);
read(fd1, str1, 80);
// Print the read string and close
printf("read: %sn", str1);
close(fd1);
}
}
此行将空字节写入 fifo:
write(fd, buffer, strlen(buffer)+1);
因此,如果管道中有两个 PID,您将读取以下字符串:
1234 2345
printf
将只打印到第一个