Question on *argv vs. *argv[]

[movl0x1]

Hi. Just wondering. Sometimes I see source code where
the *argv[] in:

ex1

int main(int argc, char *argv[])

is declared as

ex2

int main(int argc, char *argv)

I know that the ex1 is a pointer to an
array of pointers, and ex2 is a pointer to
char, but what difference is there in a C program
when the declaration is like example 2 opposed to
example 1?

thanks.

[laserlight]

More likely you saw:

int main(int argc, char **argv)

In which case it is pretty much the same thing since an array decays to a pointer to its first element.

[movl0x1]

thank you laserlight. Yes, maybe that was what I saw.

[@nthony]

technically, C distinguishes arrays from pointers (arrays have certain metadata such as sizeof an array will give you the number of its elements times the size of each element, whereas sizeof a pointer will give you just the size of that pointer). However when passed as function arguments array's lose their "arrayness" and become virtually indistinguishable from pointers. That being said, I prefer ex1 because it allows you to visualize the concept of "an array of strings" much better. Also, there's two more you may also encounter:

 int main (void)
int main (int argc, char *argv[], char *env[])

Oh, and how could I forget...

void main ()

(joking, never use that one)