[openbeos] Re: Largely Offtopic C/C++ question about pointers
- From: "Bruno van Dooren" <bruno_van_dooren@xxxxxxxxxxx>
- To: <openbeos@xxxxxxxxxxxxx>
- Date: Thu, 28 Jul 2005 20:54:40 +0200
Hi,
thanks for your reply. it was very clear.
i have gotten a reply from someone else too, and he pointed me to 2 very
useful websites:
what i have learned today: arrays and pointers might behave the same way in
lots of situations, but they must not always be treated the same.
http://pw2.netcom.com/~tjensen/ptr/pointers.htm
http://www.eskimo.com/~scs/C-faq/s6.html
others might find them useful too.
kind regards,
Bruno.
----- Original Message -----
From: <OpenBeOS.20.magnum@xxxxxxxxxxxxxxx>
To: <openbeos@xxxxxxxxxxxxx>
Sent: Friday, July 29, 2005 3:44 AM
Subject: [openbeos] Re: Largely Offtopic C/C++ question about pointers
I'm on the digest list, so maybe someone's answered this better than me.
Re example 1:
The first printf shows the same value for 'element' and '*element' because
the compiler knows that 'element' points to a t_Array. The raw value
contained in the memory allocated on the stack for 'element' is the memory
address of the beginning of 'array', so this is the output when printing
'element'. But when you dereference 'element' you get an array, and when
passing arrays to functions C/C++ passes the address of the array. This
is why printf prints the same values for 'element' and '*element'.
Note that if you printed out &array you would get the same value.
'array', '&array', and '&array[0]' are all equivalent.
The second printf prints the result you expect for array[9] (i.e. 321),
the same thing for (*element)[9] because 'element' points to 'array', but
'element' itself is just a single pointer, so if you dereference the
memory 9 spots after what it points at (which is what 'element[9]' does)
you get undefined data.
Re example 2:
The function fails to compile because to the compiler, Array is a pointer,
not an array. Taking the pointer of Array returns a pointer to the
'Array' argument to the function, not the address of the array itself.
This time, you get different values if you printf("Array = %p, &Array =
%p\n", Array, &Array);
Change the parameter to a reference and it works, i.e.
void function(t_Array& Array)
{
t_Array *ptrArray[] = {&Array}; //this works
}
Re example 3:
Like I said before, &Array within the function is a pointer to an integer,
so Array and &Array are different, whereas they are the same in the scope
where Array is actually allocated. Change it to a reference and it works
as you expect.
Note that when you declare an array, say 'int array[10]', space for 10
integers is allocated on the stack. 'array' is treated as a pointer to
the start of the array, but no space is allocated for any pointer. So
&array is kind of meaningless, and it's just the same value as 'array'.
If you declare a pointer, space is allocated on the stack for the pointer,
so 'pointer' and '&pointer' are both meaningful, and different. When you
pass an array to a function, you are passing the pointer to the start of
the array to the function as a parameter which gets put on the stack. So
taking the address of the array within the function takes the address of
the parameter on the stack.
HTH
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