How is a Hex Value manipulated bitwise?

Question

I have a very basic understanding of bitwise operators. I am at a loss to understand how the value is assigned however. If someone can point me in the right direction I would be very grateful.

My Hex Address: 0xE0074000

The Decimal value: 3758571520

The Binary Value: 11100000000001110100000000000000

I am trying to program a simple Micro Controller and use the Register access Class in the Microsoft .Net Micro Framework to make the Controller do what I want it to do.

Register T2IR = new Register(0xE0074000);
T2IR.Write(1 << 22);

In my above example, how are the bits in the Binary representation moved? I don’t understand how the management of bits is assigned to the address in Binary form.

If someone can point me in the right direction I would be very greatfull.

Answer 1

Refer to @BerggreenDK's answer for what a shift is. Here's some info about what it's like in hex (same thing, just different representation):

Shifting is a very simple concept to understand. The register is of a fixed size, and whatever bits that won't fit falls off the end. So, take this example:

int num = 0xffff << 16;

Your variable in hex would now be 0xffff0000. Note how the the right end is filled with zeros. Now, let's shift it again.

num = num << 8;
num = num >> 8;

num is now 0x00ff0000. You don't get your old bits back. The same applies to right shifts as well.

Trick: Left shifting by 1 is like multiplying the number by 2, and right shifting by 1 is like integer dividing everything by 2.