Linux Search Pattern Question:
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What is the output of this program?

#include<stdio.h>
#include<fcntl.h>

int main()
{
int fd, fd2, ret;
fd = open("san.c",O_RDONLY);
ret = close(fd2);
printf("%dn",ret);
}
a) 0
b) 1
c) -1
d) none of the mentioned

Answer:

c) -1
Explanation:
The "close" system call closes a file descriptor but in the program "fd2″ in not a file descriptor. Hence close system call returns -1.
Output:
[root@localhost google]# gcc -o san san.c
[root@localhost google]# ./san
-1
[root@localhost google]#

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What will be printed for the command below?
$ grep -c "^echo" abc
a) The count of lines that do not contain the pattern echo in file abc
b) The count of lines which begin with the pattern echo in file abc
c) The count of lines that ends with the pattern echo in file abc
d) None of the above		What is the output of this program?

#include<stdio.h>
#include<stdlib.h>
#include<fcntl.h>

int main()
{
int fd, new_fd;
char *buff;
buff = (char *)malloc(sizeof(char)*8);
fd = open("san.c",O_RDONLY);
new_fd = dup(fd);
close(fd);
read(new_fd,buff,8);
printf("%sn",buff);
}
a) this program will not print anything
b) this program will print "#include"
c) this program will give the segmentation fault
d) this program will give the syntax error