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我不明白我做错了什么，请帮助我指出。我只是在这里输入我正在尝试在程序中执行的操作，以便于理解。我创建了 3 个命名管道并将字符串写入"Hello, world!"第一个命名管道myfifo.example。现在我正在读取相同的命名管道并尝试将数据复制到第二个命名管道cmyfifo11。这种读写不会发生。它甚至没有打印线(1)和(2)。任何机构都可以纠正我。

#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#include<fcntl.h>
#include<string.h>

int main()
{
    int n, fd1, fd2;
    int pid,status;
    char    line[30]="Hello, world!\n";
    if (mkfifo("myfifo.example", 0660)<0)
        perror("Cannot create fifo");

    if (mkfifo("cmyfifo11", 0660)<0)
        perror("Cannot create fifo1");
    if (mkfifo("cmyfifo22", 0660)<0)
        perror("Cannot create fifo2");
    if((fd1= open("myfifo.example", O_RDWR))<0)
        perror("Cannot open fifo to write");  

    if ( (pid = fork()) < 0)
        perror("fork error");
    else if(pid==0){
        int z=write(fd1,line,strlen(line));
        printf("Write is done on myfifo.example\n");
        printf("CHILD PROCESS 1\n");
        fd2=open("cmyfifo11",O_RDWR);
        printf("Value of fd2=%d\n",fd2);
        if(fd2<0)
            printf("Cannot open cmyfifo11\n");
        printf("Reading\n");
        if((n=read(fd1,line,z))<0) /* Read and write is not happening */
            perror("Read error");
        printf("Value of n:%d with line %s\n",n);--->(1)
            int x=write(fd2,line,n);-------------->(2)
            printf("%d\n",x);

    }

    else if(pid>0){ printf("Parent area with %d\n",getpid());sleep(300);}
    printf("Common area\n");

    return 0;
}

输出是

Write is done on myfifo.example
CHILD PROCESS 1
Value of fd2=4
Reading
Parent area with 349
4

1 回答 1

1

您有分段错误，因为您忘记传递line给printf()

printf("Value of n:%d with line %s\n",n)

应该

printf("Value of n:%d with line %s\n",n, line);
于 2013-05-20T11:50:48.507 回答