mdh wrote On 11/29/07 17:13,:
> Hi all,
> I have gone through the FAQ and done searches in the Comp.Lang and if
> I have missed it please let me have the ref.
>
> (Question generated in part by p119).
>
>
> Given
>
> char a[10];
> char *b[8];
>
> int foo(char *s, char *[]);
>
> Now, if in main, I do this:
>
> int i;
> i=foo(a, b);
>
> it is my understanding that what is passed to "foo", as parameters,
> is
>
> 1) a pointer to [ index 0] of type "pointer-to-char" (for a)
> 2) a pointer to index 0 of type "pointer to pointer to char" ( for
> b)
Right. You may find it helpful to imagine the call
as having been written `foo(&a[0], &b[0])', which means
exactly the same thing as `foo(a, b)'.
> My question is this. If this is indeed correct, then from the
> standpoint of the function foo, what is the difference in the two
> pointers?
They point to different places (the beginning of `a'
and the beginning of `b'). Also, they point to different
kinds of things: one points to a `char' and one to a `char*'.
If you find "pointer to pointer" a confusing idea, just
think of it as "pointer to a Thing, where the Thing happens
to be a pointer to `char'".
> From the programmer's standpoint, the difference is more
> obvious. In the first case, *ptr yields a character, and in the second
> place, *ptr yields a pointer to character. I am not sure if I am
> framing this correctly.
> Perhaps, I can ask this another way. Not knowing anything about how
> the pointers were generated, if you were to be handed each pointer at
> the level of the function, could you tell the difference. If you can
> then my question is somewhat answered, if not then how does the
> compiler know what to do with each pointer.
Every pointer (in fact, every C expression) has a type,
and that type is determined at compile time and remains
fixed throughout the life of the program. An expression
that produces a `double' always produces a `double' and
never an `int' or a `struct midget'.
The type of a pointer is determined when you declare
it: If you tell the compiler that `p' is a `double*', the
compiler remembers what it was told. The compiler can
also reason about the types of expressions involving `p':
it knows that `*p' is a `double', that `p + 1' is a
`double*', that `&p' is a `double**', and so on. It works
this all out by starting from what you told it and applying
various rules.
If you tell the compiler that `b' is an array of `char*',
the same sort of thing goes on, except that the choice of
which rules to apply is a little different. Most of the
time, when you mention the name of an array you get a pointer
to the array's initial element -- hence the equivalence of
`b' and `&b[0]' in the function call. Since `b' is an
array of `char*', `b[0]' is one of those `char*' elements,
and `&b[0]' is therefore a `char**'.
When you write the function foo, part of your task is
to declare the types of the arguments it expects -- and when
you wrote `int foo(char *s, char *[]);' you communicated
this information to the compiler. The compiler remembers,
and knows that it must always provide one `char*' and one
`char**' as arguments when it calls foo. If you provide
argument expressions that aren't of these types (or can't
be converted to them automatically), the compiler complains.
Inside foo itself, the types of the function parameters
are whatever you said they were. Again, you tell the compiler
something, and it's the compiler's job to remember.
I hope this helps.
--
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