Re: fork/execv - how to get errno to the parent

On 2006-02-24, Kasper Dupont <15596462440848483280@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Jordan Abel wrote:

On 2005-09-22, Kasper Dupont <kasperd@xxxxxxxxxxx> wrote:
What you must be aware of is that some of the exec functions
may actually call /bin/sh

System does this. none of the exec functions do.

What distribution are you using? I have been able to reproduce
this on Fedora Core, Source Mage, and FreeBSD. This is the
source code I used:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main()
{
char *c="foobar";
execlp(c,c,NULL);
perror(c);
return EXIT_FAILURE;
}

foobar looks like this:
echo "Hello World"

This file is not an executable file. It is a shell script. #!/bin/sh is
missing. (these facts are both important for reasons that i explain
below)

And here is the result from stracing the program:

You have mistaken an entirely different issue for what you claimed.

It is not using an exec function that searches the path that causes
/bin/sh to be executed - it is executing a shell script that doesn't
begin with a #! line.

No. the search is done in the library.

The execlp and execvp library functions are required to search the PATH.

And that is done within the library, not with a call to /bin/sh. If the
resulting execve _fails_ with the errno ENOEXEC (meaning the file was
present and executable but not in a proper executable format), it will
then try again to call /bin/sh to interpret the given file as a shell
script.
.

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