Thread: Need help understanding bit operation

Here is a line I am confused about:

Code:

*B_CMD = SRCENX | UPDA1 | UPDA2 | PATDSEL | BCOMPEN | BKGWREN | DSTEN;

I understand that the left side is a pointer assignment. I understand that the right side is some sort of bitwise OR operation. I do not understand what it is meant to assign to the pointer. This has me totally confused. To me it seems like its ORing every bits in multiple values and then assigning end result to the pointer.

I just need the general understanding. However if someone needs the rest of the code and the specific values assigned to them I can pop that up here if needed.

Here is the rest of that function

Code:

*A1_BASE = (uint32_t)jagscreen;
*A1_FLAGS = PIXEL8 | XADDPIX | WID320 | PITCH1;
*A1_PIXEL = (y << 16) | x;
*A1_STEP = 0x010000 | (uint16_t)-8;
 
*A2_BASE = (uint32_t)(textfont + (ch << 3));
*A2_FLAGS = PIXEL1 | XADDPIX | WID8 | PITCH1;
*A2_PIXEL = 1;
*A2_STEP = 0x010000 | (uint16_t)-1;
 
*B_COUNT = 0x080008;
*B_PATD = 1;
*B_DSTD = 0;
*B_CMD = SRCENX | UPDA1 | UPDA2 | PATDSEL | BCOMPEN | BKGWREN | DSTEN;

Originally Posted by A34Chris

In this case it is the new memory location to point at? *pointer means memory address to point to and &pointer means value stored at that location if my memory is correct.

I think your memory is failing you here. There's a few different aspects of * and & when working with pointers. First, let's start with defitions

& is the address-of operator. It gives you the address of the object on it's right. It only makes sense to apply this to a variable or function, which have actual addresses in memory. If you use & on a variable of type foo, the result is of type "pointer to foo".

* is the dereference operator. It gives you the contents of the memory location on it's right. If you use this on a variable of type "pointer to foo", the result is of type "foo".

* is also use in declarations to signify that variable is a pointer to the aforementioned type.

Declarations

Code:

// a and c are pointers to int, b is a plain int -- notice that the * applies only to an individual variable, not the whole line.
// It doesn't make sense to use & in a declaration.
int *a, b, *c;

// a is a pointer, pointing to (storing the address of) b
a = &b;
// we store 42 in the memory area pointed to by a -- i.e. we store 42 in b
*a = 42;

// this makes no sense and is not valid C -- you can't dereference something that isn't a pointer
*b = 42;
// c doesn't point anywhere, so you're trying to write (assign) to invalid memory -- this is undefined behavior
*c = b;
// c is of type pointer to int; so it a, thus &a is of type pointer to pointer to int -- those types are not compatible
c = &a;
// again, c and a are of type pointer to int; *a is of type int -- those types are also not compatible
c = *a;

EDIT:
It has nothing to do with pointers or the like, but & is also the bitwise AND operator and * is also the multiplication operator. However, in these contexts, they are binary operators, requiring two operands (x & y; v *w). In the pointer context, they are unary operators, having only one operand, which is on their right.

Nevermind. Stupid question.

Thanks everyone for the help!