[openbeos] Re: Largely Offtopic C/C++ question about pointers
- From: OpenBeOS.20.magnum@xxxxxxxxxxxxxxx
- To: openbeos@xxxxxxxxxxxxx
- Date: Thu, 28 Jul 2005 20:44:11 EST
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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