I have a problem with the WIFSIGNALED()/WTERMSIG() macros, after using waitpid() - c

In this code C i launch a program from the command line and when it is closed from a signal different from SIGTERM (signal for normal end) my code should relaunch the initial program passed from the command line. But it is not so, in fact my code never relaunchs program saying that it is properly terminated.In practice my condition"if(WTERMSIG(status)!=SIGTERM)" works bad, WHY????? :'(
This is my code:
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdio.h>
int main(int argc, char*argv[])
{
pid_t pid;
int* status=(int*)malloc(sizeof(int));
int term;
if(argc<2)
{
printf("Error: Too few arguments\n");
exit(EXIT_FAILURE);
}
while(1)
{
pid=fork();
if(pid!=0) /*father*/
{
waitpid(pid,status,0);
//term=WIFSIGNALED(status);
if(WIFSIGNALED(status))
{
if(WTERMSIG(status)!=SIGTERM)
{
printf("The program %d ended abnormally:\nRelaunching...\n",pid);
sleep(1);
}
else
printf("The program %d is properly terminated...\n",pid);
break;
}
else
{
printf("Can not read the reason for termination\n");
}
}
else /*child*/
{
execvp(argv[1],argv+1);
exit(EXIT_SUCCESS);
}
}
return 1;
}

The WIFSIGNALED() and WTERMSIG() macros both expect plain ints, not pointers to ints. This means that in your code, where status is a pointer to an int, you need to use *status when calling the macros, to pass them the value of the integer.
That said: why are you calling malloc() to allocate room for a single int, anyway? Just use a normal variable, and &status if you need a pointer to it.
Also, you should return EXIT_SUCCESS from main() on successful completion of your program, not 1.

Related

Why synchronization with pipe semaphore behaves in this strange way?

I'm trying to solve some simple example of process synchronization; in this particular example I want three processes: one prints "A", another prints "B" and the last prints "C". I want to print "A" first, then "B" or "C" and again "A", "B" or "C". To be more clear I want an output pattern like "ABACACABACABABACA...."
Below there is my symple code, termination is not managed because this is just an example to understand how things work.
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
void semaphore_init (int* sw){
if (pipe(sw) == -1) {
printf ("Error\n");
exit (-1);
}
}
void semaphore_wait (int* sw){
char buffer;
if (read(sw[0],&buffer,1) != 1) {
printf ("Error\n");
exit (-1);
}
}
void semaphore_signal (int* sw){
if (write(sw[1],"X",1) != 1) {
printf ("Error\n");
exit (-1);
}
}
int s1[2];
int s2[2];
void childA();
void childB();
void childC();
int main(){
pid_t pid;
semaphore_init(s1);
semaphore_init(s2);
semaphore_signal(s1);
pid=fork();
if(pid==0) childA();
else{
pid=fork();
if (pid==0) childB();
else {
pid=fork();
if(pid==0) childC();
}
}
wait(NULL);
return 0;
}
void childA(){
while(1){
semaphore_wait(s1);
printf("A-");
semaphore_signal(s2);
}
exit(0);
}
void childB(){
while(1){
semaphore_wait(s2);
printf("B-");
semaphore_signal(s1);
}
exit(0);
}
void childC(){
while(1){
semaphore_wait(s2);
printf("C-");
semaphore_signal(s1);
}
exit(0);
}
I don't understand why the output of this code is something like:
A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B
If, instead, I put the "\n" after the char I get the right output.
Could someone, kindly, explain me the reason of that?
In many implementations, including Linux, stdout is line buffered. This means any printf characters are stored to a (memory) buffer until a newline is encountered or the line buffer fills or the stream is flushed. In the example code this results in each process writing to their own local buffer every time it gains the semaphore. So even though the sequencing is correct as you intended but because there are three seperate buffers the final output result is not as you expect. Each process eventually flushes their buffer to stdout but each of those buffers only contain the same letter repeatedly.
The solution is to either add a newline to the printf (as you have found) or to do an explicit fflush(stdout). Another alternative is to use stderr which is not buffered by default: e.g. fprintf(stderr, "A-");

C fork program explanation

I have a code as below
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
printf("a\n");
fork();
printf("b\n");
if(fork() == 0) {
printf("c\n");
exit(0);
}
printf("d\n");
return 0;
}
Output :
a
b
d
a
b
c
a
b
d
a
b
c
I don't know why the output duplicated many times.
I don't know why the output duplicated many times
Because printf() is buffered.
When a process calls fork(), the resulting child process obtains a copy of the parent's output buffer.
You can empty this output buffer by placing fflush(stdout) just before each call to fork(). In that case output should be:
a
b
b
d
c
d
c
Note that if the output refered to a terminal, it would be line-buffered by default, i.e.: the buffer would be dumped every time a \n is sent to the output. Not so if you are redirecting the output to a file.
When you call fork() it gets a copy of output buffer of the calling process. Buffering is enabled by default, so you get this behavior.
You can use
fflush(stdout);
before a call to fork(). Or, you can also disable buffering using
setbuf(stdout, NULL);
You can read more about fork here. Let me know if you need any more help.
The answer is already in the comments. You are calling fork() twice. So the solution is to just call it once and save the result in a variable like this int pid = fork(). Also, you should check if the fork-call failed (if it does, it returns a negative value).
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
printf("a\n");
int pid = fork();
if (pid < 0)
{
fprintf(stderr, "Can't fork!");
exit(1);
}
printf("b\n");
if(pid == 0)
{
printf("c\n");
}
else
{
printf("d\n");
}
return 0;
}

execl() works on one of my code, but doesn't work on another

I already used execl() in code, and it worked well.
But this time, I really have no idea why it doesn't work.
So here's the code that do not work
#include <unistd.h>
#include <stdio.h>
int main()
{
int i = 896;
printf("please\n");
execl("home/ubuntu/server/LC/admin/admin", (char*)i, NULL);
printf("i have no idea why\n");
return 0;
}
And here's the admin.c
#include <stdio.h>
#include <stdlib.h>
#include <sys/msg.h>
#include <string.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
int mid = argv[0];
printf("hi from child\n");
printf("%d\n", mid);
return 0;
}
Of course I compiled admin.c to admin, and the path isn't wrong.
>ls
admin admin.c why why.c
>pwd
/home/ubuntu/server/LC/admin
>./admin
hi from child
-1180858374
>./why
please
i have no ida why
Anyone know why it doesn't work?
My C is a bit rusty but your code made many rookie mistakes.
execl will replace the current process if it succeeds. So the last line ("i have no idea why") won't print if the child can launch successfully. Which means...
execl failed and you didn't check for it! Hint: check the typecast to char *.
You cast an int to a char * in the execl call, then again when you launch the child (admin). This is a big no-no in C. It freely allows you to misinterpret types. The only warning is most often a crash. GGC will warn you about it. I don't know about the compiler on AWS.
Check your array's bound! You don't know how many parameters admin was launched with. argv[0] always exist because it contains a representation of the program name. argv[1] may not be defined. Accessing array out-of-bound is an undefined behavior and highly dangerous.
The standard way to start another process in C is to fork the parent, then call one of the functions in the exec family to start the other process.
Consider this instead (I took the liberty to emit different messages to make them clearer).
parent.c
#include <unistd.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>
int main()
{
int i = 896;
char str[15];
int pid;
printf("Hello from parent\n");
sprintf(str, "%d", i); // convert the number into string
pid = fork();
if (pid == -1)
{
printf("Fork failed\n");
}
else if (pid == 0)
{
printf("Continue from parent\n");
}
else
{
// start the child process
execl("home/ubuntu/server/LC/admin/admin", str, NULL);
// check if it started properly
if (errno != 0)
{
printf("Error launching child process: %s\n", strerror(errno));
return 1;
}
}
printf("Goodbye from parent\n");
return 0;
}
admin.c
#include <stdio.h>
#include <stdlib.h>
#include <sys/msg.h>
#include <string.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
char * mid;
// argc is always 1 or more
if (argc >= 2)
mid = argv[1];
else
mid = "<nothing>";
printf("hello from child\n");
printf("argc = %d, argv[1] = %s\n", argc, mid);
return 0;
}

waitpid() not waiting for child

I wrote a really basic shell and for some reason, when I use fork() and then waitpid() the parent process won't wait for the child.
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <stdlib.h>
#include <sys/syscall.h>
#include <linux/limits.h>
#include "LineParser.h"
#include <termios.h>
#define MAX_STR 2048
void execute(cmdLine *pCmdLine);
int main()
{
char isContinuing = 1;
char path[PATH_MAX];
char str[MAX_STR];
char something[MAX_STR+PATH_MAX];
cmdLine* cmd;
while(isContinuing)
{
getcwd(path, PATH_MAX);
printf("%s$ ", path);
fgets(str, MAX_STR, stdin);
if(!strncmp(str, "quit", strlen("quit")))
{
isContinuing = 0;
}
else
{
cmd = parseCmdLines(str);
if(cmd->arguments != '\0')
{
execute(cmd);
}
}
}
freeCmdLines(cmd);
return 0;
}
void execute(cmdLine *pCmdLine)
{
pid_t id = fork();
if(id == 0)
{
printf("I AM CHILD.\n");
if(!execvp(pCmdLine->arguments[0], pCmdLine->arguments))
{
perror("execvp failed.\n");
exit(1);
}
exit(0);
}
printf("I AM PARENT.\n");
printf("WAITING FOR CHILD.\n");
waitpid(id);
printf("DONE WAITING\n");
}
LineParser header file is mine and it is fully working.
Now, for some reason, only the first command is working as expected,
let's assume an input "echo hi", the output is:
I AM PARENT.
WAITING FOR CHILD.
I AM CHILD.
DONE WAITING.
as expected and then it prints "hi" and the path, waiting for a command again.
For some reason, when I enter the SAME input "echo hi" the second time, the output is:
I AM PARENT.
WAITING FOR CHILD.
DONE WAITING.
$PATH$ //(WITHOUT WAITING FOR INPUT !!!)
I AM CHILD.
hi
//and here waiting for input//
Why does this happen?
There are several problems with your code:
not clearing malloc'd memory on every iteration through the while loop
putting a exit() statement in unreachable code
incorrect parameter list for the waitpid() function
unclear delination between parent code and child code in execute function
unused variable something
failed to check return value from fgets function
missing #include for sys/types.h
missing #include for sys/wait.h
IMO: the question should have included the definition of struct cmdLine
So here is a compilable version of your code. The compiler found many problems with the original code.
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <stdlib.h>
#include <sys/syscall.h>
#include <linux/limits.h>
//#include "LineParser.h"
#include <termios.h>
#include <sys/types.h>
#include <sys/wait.h> // prototype for waitpid()
//note: pid_t waitpid(pid_t pid, int *status, int options);
struct cmdLine
{
char ** arguments; // arguments[x] = ptr to an argument string
};
#define MAX_STR (2048)
#define MAX_PATH (256)
void execute(struct cmdLine *);
struct cmdLine * parseCmdLines( char * );
void freeCmdLines( struct cmdLine * );
int main()
{
char path[PATH_MAX];
char str[MAX_STR];
//char something[MAX_STR+PATH_MAX];
struct cmdLine* pCmd = NULL;
while(1)
{
getcwd(path, PATH_MAX);
printf("%s$ ", path);
if( NULL == fgets(str, MAX_STR, stdin) )
{
perror( "fgets failed" );
exit( EXIT_FAILURE );
}
// implied else
if(!strncmp(str, "quit", strlen("quit")))
{ // then strings equal
break; // exit while loop (and pgm)
}
// implied else input not equal 'quit'
pCmd = parseCmdLines(str);
if( (NULL != pCmd) && (NULL != pCmd->arguments) )
{ // then one or more arguments entered/parsed
execute(pCmd);
} // end if
freeCmdLines(pCmd); // free all strings memory, then free struct memory
pCmd = NULL; // cleanup
} // end while
return 0;
} // end function: main
void execute(struct cmdLine *pCmdLine)
{
int status = 0;
pid_t id = fork();
if(id == 0)
{ // then, child
printf("I AM CHILD.\n");
if(!execvp(pCmdLine->arguments[0], pCmdLine->arguments))
{ // if no error then never gets here
perror("execvp failed.\n");
} // end if
}
else
{ // else, parent
printf("I AM PARENT.\n");
printf("WAITING FOR CHILD.\n");
waitpid(id, &status, 0);
printf("DONE WAITING\n");
} // end if
} // end function: execute
You invoke undefined behavior by calling the waitpid() function with the wrong number of arguments. Anything could happen.
This simplified variant of your code works fine for me:
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
int main ()
{
int i;
for (i = 0; i < 3; i += 1)
{
pid_t id = fork();
if(id == 0)
{
char *argv[] = { "echo", "hi", NULL };
printf("I AM CHILD.\n");
execvp("echo", argv);
/* failed to exec */
perror("execvp failed.\n");
exit(1);
} else if (id < 0) {
perror("fork failed.\n");
exit(1);
}
printf("I AM PARENT.\n");
printf("WAITING FOR CHILD.\n");
waitpid(id, NULL, 0);
printf("DONE WAITING\n");
}
return 0;
}
Your call to waitpid(2) is wrong.
According to man 2 waitpid, it's:
pid_t waitpid(pid_t pid, int *status, int options);
You probably need to define an int and call it as:
waitpid(id, &status, 0);
or use the simpler version wait(2), which will work for any child:
wait(&status);
Your main problem is that you don’t let the compiler check your code. You should generally enable the compiler warnings and try to understand them.
$ gcc -Wall -Wextra -Werror -Os -c myshell.c
This is the minimum command line I use. When your code compiles with these settings, you have already eliminated a bunch of hard-to-find bugs in your code. Among these bugs is, as others already have mentioned, the call to waitpid.
Have a look at http://pubs.opengroup.org/onlinepubs/7908799/xsh/waitpid.html. The Open Group specification requires that you #include the two headers <sys/types.h> and <sys/wait.h> before using the waitpid function. Your program doesn’t do this.

Using fork() for split process- what's wrong with the program

I'd really love your help with understanding why doesn't the process reach the "son process" after using fork() command. I'm trying to write a program that runs another program, but It seems that the program dosen't even reach the son process. I can tell that since "son process" is not being printed to the screen, and I really wonder why.
Here's a sketch of the code- I can't even check if it is alright since as I said, it doesn't even reaching the son process, I always get "son exited with error".
#include <sys/types.h>
#include <sys/wait.h>
#include <stdlib.h>
#include <assert.h>
#include <signal.h>
#include <string.h>
#include <stdio.h>
#include <dirent.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <fcntl.h>
#include <errno.h>
#include <string.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/time.h>
#include <time.h>
#define MAXARGV 5;
int main() {
char* cmd;
int child_status;
char* s;
char** argv;
int counter;
cmd= (char*) calloc( 5, sizeof(char)*20);
s=(char*) calloc(1,sizeof(char)*20);
argv=(char**) calloc(5, sizeof(char*)*20);
printf("Please write a command\n");
gets(cmd);
counter = 0;
while (strcmp(cmd, "exit") != 0) {
int pid = fork();
if (pid == 0) {
printf("son process");
while (sscanf(cmd, "%s", s) == 1) {
strcpy(argv[counter], s);
counter++;
}
execv(argv[0], argv);
printf("the command is not legal");
assert(0);
}
else {
if (wait(&child_status) == -1) {
printf("error waiting for pid=%d\n", pid);
exit(-1);
}
if(WIFEXITED(child_status)!=0)
printf("son status=%d\n", WEXITSTATUS(child_status));
else
printf("son exited with error\n");
}
printf("Please write a command");
gets(cmd);
}
free(s);
free(cmd);
free(argv);
printf("here as well");
return 1;
}
The program reaches the printf("son process") just fine, but that just puts the string in a buffer inside the process and since you didn't fflush() it, it doesn't make it to the screen and is discarded with the rest of the process' memory in the exec call. Note, that stdout is normally line-buffered, so if you had newline there, it would auto-flush. Also stderr is by default unbuffered and more suitable for debug prints (fprintf(stderr, "child process")).
You are trying to assemble the command read from standard input in argv, but it only has memory for the actual arguments given to you, so you overrun this memory and get segmentation fault.
if WIFEXITED gives zero, you should use WIFSIGNALED and WTERMSIG to confirm that the error is indeed SIGSEGV.
assert(0) is not a good way to terminate process after error. exit(1) is. Assertions are only for conditions that indicate bug in the code itself if they happen and are often eliminated (by defining NDEBUG) from production code.

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