printing agruments using text interpreter and execl - c

Question

I am trying to print command line arguments using execl function which executes a text interpreter but I am not able to get the first argument printed.
this is my main program
#include<unistd.h>
#include<stdio.h>
#include<fcntl.h>
#include<sys/types.h>
#include<sys/stat.h>
int main(int argc,char *argv[])
{
pid_t pid;
if((pid=fork())<0)
printf("error\n");
else if(pid==0)
if((execl("textinterpreter","this","is","usp","lab",(char*)0))<0)
perror("error in execl\n");
if(waitpid(pid,NULL,0) !=pid)
printf("error1\n");
system("ls > list");
return 0;
}
This is my textinterpreter file
#!/home/chirag/echoarg1 my1
This is my echoarg1.c file
#include<stdio.h>
#include<unistd.h>
#include<fcntl.h>
#include<sys/types.h>
#include<sys/stat.h>
main(int argc,char *argv[])
{
int i;
for(i=0;i<argc;i++)
printf("argv[%d]=%s\n",i,argv[i]);
}
The output I am getting is
argv[0]=/home/chirag/echoarg1
argv[1]=my1
argv[2]=textinterpreter
argv[3]=is
argv[4]=usp
argv[5]=lab
Where as the expected output is
argv[0]=/home/chirag/echoarg1
argv[1]=my1
argv[2]=textinterpreter
argv[3]=this
argv[4]=is
argv[5]=usp
argv[6]=lab
Can someone please point out the mistake.

Answer 1

from the execl man page:
int
execl(const char *path, const char arg0, ... /, (char *)0 */);
and
The initial argument for these functions is the pathname of a file which
is to be executed.
The const char *arg0 and subsequent ellipses in the execl(), execlp(),
and execle() functions can be thought of as arg0, arg1, ..., argn.
Together they describe a list of one or more pointers to null-terminated
strings that represent the argument list available to the executed program. The first argument, by convention, should point to the file name
associated with the file being executed. The list of arguments must be
terminated by a NULL pointer.
so in your line:
if((execl("textinterpreter","this","is","usp","lab",(char*)0))<0)
"textinterpreter" is the path of the file to execute, but execl is expecting the next argument, arg0, to "point to the file name associated with the file being executed".
in other words it should look more like:
if((execl("textinterpreter","textinterpreter","this","is","usp","lab",(char*)0))<0)
Usually the first two args to execl would likely be the same, but it allows you the flexibility to specify one arg for finding the file to execute and another argument to pass as argv[0].
As an experiment, try changing your execl to call echoarg1 directly like this:
if((execl("echoarg1","FAKEPATH","this","is","usp","lab",(char*)0))<0)
You should see output like:
argv[0]=FAKEPATH
argv[1]=this
argv[2]=is
argv[3]=usp
argv[4]=lab
In your example it appears that it's displaying "this" instead of "textinterpreter" as a result of the how the bash interpreter handles your "textinterpreter" script. The bash interpreter appears to acting as follows:
takes the hash-bang line and runs /home/chirag/echoarg1
pass it argv[0]=/home/chirag/echoarg1
pass it the rest of the hashbang line as additional args (in this case just argv[1]=my1)
as it appends the original argv list it appears use the executable path ("textinterpreter") instead of the actual argv[0] ("this") and then continues with argv[1] ("is") and the rest...

Answer 2

It is not the bash interpreter (whatever that's supposed to mean - there is no bash involved here) which discards arg0, but it is the kernel's function load_script() with the statement
retval = remove_arg_zero(bprm);
above it there is a comment saying
* Splice in (1) the interpreter's name for argv[0]
* (2) (optional) argument to interpreter
* (3) filename of shell script (replace argv[0])
- in your case, (1) /home/chirag/echoarg1, (2) my1, (3) textinterpreter (replace this).