Execl dont start the program

Asked Jun 19 '19 at 12:28

Active Jun 19 '19 at 12:47

Viewed 58 times

I have two programs, one of them calls another through execl, but it doesn't start.

Here is the calling program:

#include <stdio.h>
#include <unistd.h>
int main(int argc, int *argv[])
{
  printf("Program %s executes", argv[0]);
  execl("hello", " ", "Hello", "World!", NULL);
  return 0;
}

The "hello" program:

#include <stdio.h>
int main(int argc, char *argv[])
{
  int i=1;
  printf("%s\n",argv[0]);
  printf("Program started and got string : ");
  while(argv[i++] != NULL)
   printf("%s ",argv[i]);
  return 0; 
}

edited Jun 19 '19 at 12:47

asked Jun 19 '19 at 12:28

Tomas

2

Possible duplicate of [How to call execl() in C with the proper arguments?](https://stackoverflow.com/questions/12596839/how-to-call-execl-in-c-with-the-proper-arguments) – KompjoeFriek Jun 19 '19 at 12:31
4

What does the `execl` return, and what `errno` does it set? – Angew is no longer proud of SO Jun 19 '19 at 12:31
1

Does `execl` find your hello program? maybe you need `execl("./hello", ... ) ` ? – Øystein Schønning-Johansen Jun 19 '19 at 12:33
thank you very much, he needed ./hello – Tomas Jun 19 '19 at 12:40
1

`while(argv[i++] != NULL) printf("%s ",argv[i]);` does not print the first argument and also includes the `NULL` pointer at the end. – mch Jun 19 '19 at 12:41
it does i just typed the code here wrong sorry) – Tomas Jun 19 '19 at 12:45

Execl dont start the program

0 Answers0