When I call execvp, for example execvp(echo, b) where b is an array of arguments for the command a, will changing this array later affect the execvp call made previously? When I try calling execp(echo, b), it ends up printing out (null) instead of the content inside of b. Can anyone point out why and what I have to do to pass the arguments correctly?
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2 Answers
After you call exec() or one if its relatives, your original program doesn't exist anymore. That means nothing in that program can affect anything after the exec() call, since it never runs. Maybe you're not building your arguments array correctly? Here's a quick working example of execvp():
#include <unistd.h>
int main(void)
{
char *execArgs[] = { "echo", "Hello, World!", NULL };
execvp("echo", execArgs);
return 0;
}
From the execvp() man page:
The
execv(),execvp(), andexecvpe()functions provide an array of pointers to null-terminated strings that represent the argument list available to the new program. The first argument, by convention, should point to the filename associated with the file being executed. The array of pointers must be terminated by aNULLpointer.
A common mistake is to skip the part about "The first argument, by convention, should point to the filename associated with the file being executed." That's the part that makes sure echo gets "echo" as argv[0], which presumably it depends on.
4 Comments
Lucas
Instead of using echo, I used another program that essentially prints out everything in its argv array. I always assumed argv[0] would be the name of the command itself (the first param of execv), but in this case after calling execv, argv[0] was not. Rather it was the second param of execv. Can you clarify this?
Carl Norum
Well, when you passed the argument list to
execv, did you set argv[0] to the name of the command? The array you pass as the second argument execvp becomes argv in the new program, so you need to set it up the way it would expect it to be.
user877329
This code breaks const correctness. I would rather write const char* args[]={...}; but the prototype for execvp requires the strings in args to be writable!
Celeritas
Thanks you helped me. Isn't this sort or redundant having to specify the name of the command twice? Is there ever a time the first arugment to execvp would be different to the first element of the second argument to execvp? Why isn't this done automatically and why don't compilers warn about this?
Remember, that after exec call your program is exchanged by a new one. It's not executing anymore, so any code in the same process after exec call is, in fact, unreachable.
Are you sure that b array is terminated with NULL? The last element must be NULL for exec to work correctly. Also, remember to set your first parameter to "echo" as well (it's argv[0]).
Try
execlp("echo", "echo", "something", NULL);
Btw, execlp is a bit more comfortable to use, you can pass as many parameters as you wish.
4 Comments
Lucas
From what I understand, execvp takes in the command and the argv array. At the time of the execvp call, both the command and the argv array were not NULL. However, when echo was called it didn't seem to have any arguments.
Carl Norum
But the OP is asking about
execvp. How is your answer going to help?
Piotr Zierhoffer
It only another convention of writing the same thing. All these functions are just proxies for execve, afaik.
Kaz
The macro
NULL is allowed to expand to an integral constant expression. Such an expression serves as a pointer in a pointer context. But the trailing argument of a variadic function is not such a context. You need (char *) 0 or (char *) NULL.