Named & Un-named Pipe Calls - Linux Debugging Questions & Answers

This set of Linux Debugging questions and answers focuses on Named and Un-named Pipe Calls.

1. What is the output of this program?

```
   #include<stdio.h>
```
```
   #include<string.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd[2];
```
```
       int count;
```
```
       char buffer[6];
```
```
       if( pipe(fd) != 0)
```
```
           perror("pipe");
```
```
       memset(buffer,'\0',6);
```
```
       count=write(fd[1],"Linux",6);
```
```
       read(fd[0],buffer,6);
```
```
       printf("%s\n",buffer);
```
```
       return 0;put
```
```
   }
```

a) this program will print the string “Linux”
b) this program will print nothing because the buffer is empty
c) segmentation fault
d) none of the mentioned
View Answer

Answer: a
Explanation: The data in buffer is written by the pipe.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
Linux
[[email protected] sanfoundry]#

2. What is the output of this program?

```
   #include<stdio.h>
```
```
   #include<stdlib.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd[2];
```
```
       int child;
```
```
       char buff[6];
```
```
       if(pipe(fd) != 0)               
```
```
           perror("pipe");
```
```
       child=fork();
```
```
       switch(child){
```
```
           case -1 :
```
```
               perror("fork");
```
```
               exit(1);
```
```
           case 0 :
```

               if (write(fd[1],"Linux",6) != 6)

```
                   perror("write");
```
```
                   break;
```
```
               default :
```
```
                   read(fd[0],buff,6);
```
```
                   printf("%s\n",buff);
```
```
                   break;
```
```
       }
```
```
       return 0;
```
```
   }
```

a) this program will print the string “Linux”
b) this program will print nothing
c) segmentation fault
d) none of the mentioned
View Answer

Answer: a
Explanation: One process can read and write to another process by pipe.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
Linux
[[email protected] sanfoundry]#

3. What is the output of this program?

```
   #include<stdio.h>
```
```
   #include<fcntl.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd;
```
```
       char buff[512];
```

       if( mkfifo("/tmp/test_fifo",0666) == -1)

```
           perror("mkfifo");       
```

       fd = open("/tmp/test_fifo",O_RDONLY);

```
       read(fd,buff,512);
```
```
       printf("%s\n",buff);
```
```
       return 0;
```
```
   }
```

a) this program will print the garbage of 512 bytes
b) this program will print nothing
c) segmentation fault
d) none of the mentioned
View Answer

Answer: b
Explanation: In this program the fifo is opened in read only mode. Hence the process will remain block on read.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
^Z
[12]+ Stopped ./san
[[email protected] sanfoundry]#

4. This program will print the _____ string.

```
   #include<stdio.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd[2];
```
```
       char buff[11];
```
```
       if (pipe(fd) != 0)
```
```
           perror("pipe");
```
```
       write(fd[1],"Sanfoundry",11);
```
```
       lseek(fd[0],0,3);
```
```
       read(fd[0],buff,11);
```
```
       printf("%s\n",buff);
```
```
       return 0;
```
```
   }
```

a) “Sanfoundry”
b) “San”
c) “foundry”
d) none of the mentioned
View Answer

Answer: a
Explanation: The “lseek” system call does not work with pipes.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
Sanfoundry
[[email protected] sanfoundry]#

5. What is the output of this program?

```
   #include<stdio.h>
```
```
 
```
```
   int main()
```
```
   {
```

       if (mkfifo("/tmp/test_fifo",0666) != 0)

```
           perror("mkfifo");
```

       if (mkfifo("/tmp/test_fifo",0666) != 0)

```
           perror("mkfifo");
```
```
       return 0;
```
```
   }
```

a) this program will create two named pipes “test_fifo” in the /tmp directory
b) this program will create one named pipe “test_fifo” in the /tmp directory
c) segmentation fault
d) none of the mentioned
View Answer

Answer: b
Explanation: In this program when the mkfifo executes second time, the fifo already exists. Hence it gives error.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
mkfifo: File exists
[[email protected] sanfoundry]#

6. What is the output of this program?

```
   #include<stdio.h>
```
```
   #include<fcntl.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd, count;
```
```
       char buff[10];
```

       if (mkfifo("/tmp/test_fifo",0666) != 0)

```
           perror("mkfifo");
```

       fd = open("/tmp/test_fifo",O_RDONLY|O_NONBLOCK);

```
       count = read(fd,buff,10);
```
```
       printf("%d\n",count);
```
```
       return 0;
```
```
   }
```

a) 0
b) -1
c) 10
d) none of the mentioned
View Answer

Answer: a
Explanation: The read system call will return 0 bytes.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
0
[[email protected] sanfoundry]#

7. This program will print the value

```
   #include<stdio.h>
```
```
   #include<fcntl.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int rfd, wfd, count;
```
```
       char buff[11];
```

       if (mkfifo("/tmp/test_fifo",0666) != 0)

```
           perror("mkfifo");
```

       wfd = open("/tmp/test_fifo",O_WRONLY|O_NONBLOCK);

       count = write(wfd,"Sanfoundry",11);

```
       printf("%d\n",count);
```

       rfd = open("/tmp/test_fifo",O_RDONLY|O_NONBLOCK);

```
       count = read(rfd,buff,11);
```
```
       return 0;
```
```
   }
```

a) 0
b) -1
c) 11
d) none of the mentioned
View Answer

Answer: b
Explanation: The “write” system call will return -1 because named pipe is open only for writing. In named pipe writing can only be done when it is open for read and write both.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
-1
[[email protected] sanfoundry]#

8. What is the output of this program?

```
   #include<stdio.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       int fd[3],count;
```
```
       if (pipe(fd) != 0)
```
```
           perror("pipe");
```
```
       count = write(fd[2],"Hello",6);
```
```
       printf("%d\n",count);
```
```
       return 0;
```
```
   }
```

a) 6
b) 0
c) -1
d) segmentation fault
View Answer

Answer: c
Explanation: Data can be written only in fd[1].
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
-1
[[email protected] sanfoundry]#

9. In this program the fifo “my_fifo”

```
   #include<stdio.h>
```
```
 
```
```
   int main()
```
```
   {
```
```
       if (mkfifo("my_fifo",0666) != 0)
```
```
           perror("mkfifo");
```
```
       return 0;
```
```
   }
```

a) can not be created
b) will be created in present working directory
c) will have the execute permissions
d) none of the mentioned
View Answer

Answer: b
Explanation: None.
Output:
[[email protected] sanfoundry]# ls
san san.c
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
[[email protected] sanfoundry]# ll
total 12
prw-r–r–. 1 root root 0 Aug 21 15:04 my_fifo
-rwxr-xr-x. 1 root root 4925 Aug 21 15:04 san
-rw-r–r–. 1 root root 99 Aug 21 15:01 san.c
[[email protected] sanfoundry]#

10. What is the output of this when the pipe is successfully created?

```
    #include<stdio.h>
```
```
 
```
```
    int main()
```
```
    {
```
```
        int ret_val;
```
```
        int fd[2];
```
```
        ret_val = pipe(fd);
```
```
        printf("%d\n",ret_val);
```
```
        return 0;
```
```
    }
```

a) 0
b) -1
c) 1
d) none of the mentioned
View Answer

Answer: a
Explanation: The “pipe” system call returns 0 on the successfull creation of the pipe.
Output:
[[email protected] sanfoundry]# gcc -o san san.c
[[email protected] sanfoundry]# ./san
0
[[email protected] sanfoundry]#

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