I don't understand why *argv[] and **argv have the same meaning when used as a fct parameter.
char **foo; /* a pointer to a pointer to char */
char *bar[]; /* an array of pointers to char, of unspecified size */
*argv[] is nothing more than syntactic
sugar for **argv (Note that this discussion is valid only for argv when used as function
parameter. )
An array is converted to a pointer to its first element when it is passed as an argument. A simpler example:
is equivalent to:Code:void print_numbers(int numbers[]);
Code:void print_numbers(int *numbers);
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Because they mean exactly the same thing. You can't pass arrays. It's an illusion. You always pass a pointer to the first element instead. So the two syntaxes mean the same thing, but are typed differently.
is also equivalent to:Code:void foo(int* numbers[]);
Code:void foo(int** numbers);
