hellofork.c ∗ :
/* * Simple Hello world program with 2
processes.
*/ #include <stdio.h>
main() { printf("\%d:
Hello\n", getpid());
if (
!fork() )
printf("\%d: world\n", getpid());
}
If we add a pipe to the program more (a filter to watch content sequentially one screen at a time) the output is:
terminal~> ./hellofork |
more terminal~> 53438:
Hello terminal~> 53438:
Hello terminal~> 53440:
world
Why does this happen
Related
I am using Ubuntu 18.04. I am currently doing a course on Operating Systems and was just getting used to fork() and exec() calls.
I am running the following C program
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int
main(int argc, char *argv[])
{
printf("hello world (pid:%d)\n", (int) getpid());
int rc = fork();
if (rc < 0) {
// fork failed; exit
fprintf(stderr, "fork failed\n");
exit(1);
} else if (rc == 0) {
// child (new process)
printf("hello, I am child (pid:%d)\n", (int) getpid());
} else {
// parent goes down this path (original process)
printf("hello, I am parent of %d (pid:%d)\n",
rc, (int) getpid());
}
return 0;
}
On running the code in Sublime Text Editor with the build file
{
"cmd" : ["gcc $file_name -o ${file_base_name} && ./${file_base_name}"],
"selector" : "source.c",
"shell": true,
"working_dir" : "$file_path"
}
I get the result
hello world (pid:16449)
hello, I am parent of 16450 (pid:16449)
hello world (pid:16449)
hello, I am child (pid:16450)
Whereas if I use the terminal and run the same code using gcc,
I get
hello world (pid:17531)
hello, I am parent of 17532 (pid:17531)
hello, I am child (pid:17532)
Now I know that the latter is the correct whereas the output I get in Sublime is the wrong one. How can the outputs when the compiler I am using remains the same can be different?
hello guys i am playing CTF and i have to crack a program to get shell the source code is :
/*
* gcc ch21.c -lcrypt -o ch21
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <crypt.h>
#include <sys/types.h>
#include <unistd.h>
int main (int argc, char *argv[]) {
char pid[16];
char *args[] = { "/bin/bash", "-p", 0 };
snprintf(pid, sizeof(pid), "%i", getpid());
if (argc != 2)
return 0;
printf("%s=%s",argv[1], crypt(pid, "$1$awesome"));
if (strcmp(argv[1], crypt(pid, "$1$awesome")) == 0) {
printf("WIN!\n");
execve(args[0], &args[0], NULL);
} else {
printf("Fail... :/\n");
}
return 0;
}
now i debugged it with gdb as i understood from the source i have to enter proccessid (PID) during runtime to get successful shell with GDB-PEDA i have tried getpid during breakpoint but how to continue with proccess id with gdb only run command pass input to the program any help !
any notify !
Not sure if I understood your question correctly, but PID is limited in range and cycle when there limit is reached and the max is usually around 2^15. You could simply run a loop that would run through the potential PID to match the one that will be assigned for the process.
Something like this would do:
import os, crypt, subprocess
pid = os.getpid()+50 #safe buffer for things created after python script was started
print "Selected: ",pid
for i in range(32768):
sp = subprocess.Popen(['./ch21', crypt.crypt(str(pid), "$1$awesome")], stdout=subprocess.PIPE)
output = sp.stdout.readline()
if "Fail" not in output:
print output
break
I have written a CMakeFile that simply creates a test:
cmake_minimum_required (VERSION 3.1)
project (hello_test)
enable_testing ()
set (all_tests "")
list (APPEND all_tests hello_test)
add_executable (hello_test hello_test.c)
add_test (NAME hello_test COMMAND hello_test)
The test is as follow :
#include <stdio.h>
#include <unistd.h>
int main()
{
printf ("This message should appear when the test is launched \n"); //1st message
sleep(5); //sleep for 5s
printf ("This message should appear after 5 s \n"); //2nd message
}
I run the generated test using ./hello_test.exe.
What I expected is that the 1st message appear and then the program sleep for 5s and then the 2nd message appears.
However what I see is that the messages are only shown in the end of execution of the program
So if I want to debug a concurrent test(see which instruction is executed before the other using printf) it want help me a lot.
Is there a way to change the behaviour of the program as I expected ?
The printf buffer has not been flushed you could either flush to stdout or do fprintf(stdout, "...");
#include <stdio.h>
#include <unistd.h>
int main()
{
printf ("This message should appear when the test is launched \n"); //1st message
fflush(stdout); // add this
sleep(5); //sleep for 5s
printf ("This message should appear after 5 s \n"); //2nd message
}
Another thing could be
#include <stdio.h>
#include <unistd.h>
int main()
{
fprintf (stdout, "This message should appear when the test is launched \n"); //Notice the fprintf
sleep(5); //sleep for 5s
fprintf (stdout, "This message should appear after 5 s \n"); //Notice the fprintf
}
for(i=0;i<2;i++)
if(fork()==0)
printf("Hi");
I am expecting 3 hi and getting 4 hi
SO i edited the printf as printf("Hi %d %d %d ",i,getpid(),getppid());
The first child created prints two hi with same value of I i.e 0 and its pid and parent's pid are also same. Why?
It's quite interesting and looks like the answer is output buffering. For example we have:
#include <unistd.h>
#include <stdio.h>
int main() {
for(int i=0;i<2;i++) {
if(fork()==0) {
printf("Hi %d %d %d\n",i,getpid(),getppid());
}
}
}
If run this code in terminal there will be 3 lines, but if I will redirect the output to less there will be four!
If we will flush the buffer after printf() the problem will disappear:
// ...
printf("Hi %d %d %d\n",i,getpid(),getppid());
fflush(stdout);
// ...
That's happening because stdout is buffered, so when process forked the buffer still not flushed.
From man stdout:
The stream stderr is unbuffered. The stream stdout is
line-buffered when it points to a terminal. Partial lines will not
appear until fflush(3) or exit(3) is called, or a newline is printed.
This can produce unexpected results, especially with debugging
output.
I'm trying to simulate a pipe behavior on Ubuntu's Terminal, for example the command:
"echo hello | wc".
Please assume I got the tokens from stdin, handled everything correctly and now These are the commands I "received" from the user who typed them in the shell for me to handle.
I'm trying to create two processes. Using a pipe, in the first process, I point the file descriptor of the writing edge of the pipe to stdout. The second process should read into stdin with the reading edge of the pipe what execvp(..) returned.?
Here is the code I wrote:
#include <stdio.h>
#include <stdlib.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
int main()
{
char* fcmd[] = {"echo", "hello", NULL};
char* scmd[] = {"wc", NULL};
pid_t pid;
int pipe_des[2];
int i;
pipe(pipe_des);
for(i = 0; i < 2; i++)
{
pid = fork();
if (pid ==0)
{
switch (i)
{
case 0: // FIRST CHILD
{
dup2(pipe_des[1], STDOUT_FILENO);
close(pipe_des[0]);
execvp(fcmd[0], fcmd);
exit(0);
}
case 1: //SECOND CHILD
{
dup2(pipe_des[0], STDIN_FILENO);
close(pipe_des[1]);
execvp(scmd[0], scmd);
exit(0);
}
}
}
else if (pid < 0)
exit(EXIT_FAILURE);
return EXIT_SUCCESS;
}
I get: " amirla#ubuntu:~/Desktop/os/class/ex4$ 1 1 6 "
Like it should, but why he's printing the bash cwd first? The pipe seems to work because I get what I should, according to the length of the word I'm sending with the echo command(in the main()). After that the cursor just waits on the line below for another command without showing me the bash pwd. (maybe stdin is waiting?)
I've looked in many posts on here as well as on other websites and I still can't seem to find a solution to my problem. Any help would be appreciated. Thanks in advance.
Note: Please Ignore checking for errors, I've delete them to make the code shorter so assume they exist.
Why do I get a prompt before the output?
Your main process doesn't wait for the children to finish. What you see is:
Main starts
Main creates children
Main exits
BASH prints prompt
Children start their work
To prevent this, you need to wait for the children. See How to wait until all child processes called by fork() complete?
In your case, it's enough to add
waitpid(-1, NULL, 0);
after the loop.