Re: execv fails with EFAULT

[Prasanta Sadhukhan <Prasanta.Sadhukhan@Sun.COM>]

Steve Graegert wrote:

> On 3/7/07, Prasanta Sadhukhan <Prasanta.Sadhukhan@sun.com> wrote:
>
>> Hi,
>>
>> When I tried the following program
>> void test4()
>> {
>>     printf("doing execv\n");
>>     if(execv("/bin/ls", "-l") == -1)
>>         printf("exec failed with errno %d\n", errno);
>>     else
>>         printf("exec succeed\n");
>> }
>
>
> Prasanta,
>
> A NULL terminated array of arguments must be passed to execv(2) as the
> prototype indicates: int execv(const char *path, char *const argv[]);
>
> For example:
>
> #include <unistd.h>
>
> char *cmd[] = { "ls", "-l", NULL };
> int result = execv ("/bin/ls", cmd);
>
> An error of EFAULT usually means that an argument points to an illegal 
> address.
>
>     \Steve

Thanks Steve... If I have a string like this sprintf(str, 
"-Xparameter:%d %s, value, command"), how to make it NULL terminated. Is 
this declaration char *cmd[] = {str, NULL} and invocation
execv(path, cmd) correct?

Regards
Prasanta

>
> -- 
>
> Steve Grägert <steve@graegert.com>
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