Does a PC overwrite all bits when updating memory from 1100 to 1111, even if some match?

Hi guys ! I really need your help!lets assume the current data on specific address is : 1100and whatsoever
procedure the PC does, the data of that specific address need to be updated to :1111
so here is my question, the previous data 1100 is gone .. but I want to ask, in first two digits of (1100) which they are 11 are the same as the two digits of the updated data ( the first two digits of the updated data are 11) , my question are the PC keeps the two first digits as it's and not touch them ? I mean once the PC want to updated 1111 instead of 1100, the first two digits in the previous data as the same the updated data, my question, is the PC keeps the first two digits as it's in the previous data or he actually touch them and change them from 11 to 11 ? I'm talking just on first two digits related to the previous data over the updated data .. 

For example the previous data in last two digits was 00 and in the new updated data the last two digits are 11, so the PC change them from 00 to 11 as new updated data reveals !but what about if it was 11 and and in the updated data was also 11, does the pc change from 11 to 11 as stupid?

>> I really need your help!Why is this question so important to you?

Here's the deal -- from what you describe we have a 4-bit storage location that can hold 4 binary digits, or bits:x x x xLets number these bits from 3 to 0:3 2 1 0
x x x x
Each bit can be 0 or 1Based on previous activity, the computer starts off with this location containing 11003 2 1 0x x x x
1 1 0 0

Now the computer decides to load 1111 into this location.There are two ways of doing this:a) The computer simply loads the new 4-bit 1111 value into the memory location on the basis that it really doesn't care what's already in there.
b) -- The computer compares the new bit 3 to the old bit 3 and says to itself "these are the same so I won't bother re-writing this bit"-- The computer compares the new bit 2 to the old bit 2 and says to
itself "these are the same so I won't bother re-writing this bit"-- The computer compares the new bit 1 to the old bit 1 and says to
itself "these are different so I will re-write this bit"-- The computer compares the new bit 0 to the old bit 0 and says to
itself "these are different so I will re-write this bit"
Option (a) will require a single operation and can be performed in a single clock cycle -- this the option you describe as "stupid"
Option (b) will require multiple operations and will consume multiple clock cycles (thereby taking more time and consuming more power).
Which option would you choose if you were designing a computer?

Let's think about a single bit of memory or a single register bit.Each kind of memory/register bit has a mechanism to remember its current value.  It uses as little power as possible to remember the current value.Each kind of memory/register bit has a mechanism to write a new value.  This is much stronger than the mechanism that remembers the current value, so the new value overwrites the old value.  When the write is complete, the low-power mechanism takes over and remembers the new value just written.The computer doesn't care what value was in the memory/register bit previously -- the write operation simply replaces it with the new value.  If the new value is different from the old value, it may take a little more energy to flip the memory/register bit but the difference is usually trivial.
Hope this helps.

I almost understand you, for answering your question lemme ask you one question before and hope I will afterwards answer your question.
lets assume that I have switch which at left side there was 5v, the switch is opened, and on the right side 5v, so my question when I close the switch then the voltage at two side will still 5v? I mean when I close the switch the left side have 5v so it will traverse to the right side butin right side there's also 5v .. so the closed switch is like bridge that's having 5v at every point on the switch after we close it .. am I right?!!!!
I mean by this witch : 5v------5v so when we close the switch , it would be like this 5v-----5v so at every point on the straight line after we closing the switch will be the voltage 5v?!

We've talked about this (oh, so many times) before. We measure different voltages from a common reference point -- typically 0V, which may also be referred to as "ground" or GND.
Another name for voltage is "electrical potential" -- and one way to think about it is in comparison to the potential energy associate with an object when you lift it off the ground.Suppose we have two independent tanks of water, where the height of the water in each tank is the same (5 feet in this case) as measured from a common ground level. Now, just for giggles and grins, suppose we connected these two tanks together with a channel as illustrated below:

If you were to measure the height of the water (as compared to the ground) at any point along its upper surface -- it would always be five feet.If you have two 5-volt batteries and you connect them together with a piece of conducting copper wire, then the voltage (as compared to 0V) at any point on the wire will be 5V

Ryan -- to help us better help you and answer your questions, can you please share some information with us as follows:a) How old are you?b) Are you studying electronics on your own, or are you on some sort of course? If the latter, what course and what institution?

Yes every storage device or say a register has its own mechanism to store its current value..WIth minimum power it stores the current value 

(a) 21 years old(b) study by myself.
(c) is that stupid question to ask like this? I mean ..am I stupid or not good to ask like those questions?!

I promised you to answer your question.Yes, I would choose Option(a) as far as a designer looks for descending the time's burst of doing the operation, for example lets assume you're using cellphone, and lets assume you used specific operation lets assume "delete" .. so we are struggling as designers to not let you wait till the operation of delete get finished, and you know it gets the USER to nervous if it's taking too much time!  

but still confusing me your answer to (a) , you said that the PC loads "1111" doesn't matter to what it was, so lets assume there was 1100, now I want to load "1111" , so implicitly the PC clear "1100" by grand of the capacitors , and so we get "0000" and will load "1111" ?the confusing part is, if it was 1 on specific bit, and now I want to load 1, but there's already 1 , so how 1 will be replaced by 1? is the previous one cleared first and then charged to 1? because lets assume it was 1 and I want to charge to 1 .. but it's already 1 so there's no current from 1 to 1 because we are having the same voltage .. so?! I think the PC keeps the voltage as its previous state

I know what's register .. but may you please attach a diagram of your explanation? I'm not understanding that much.so if I was having 1 on register, and I want to load 1 to register(in other words I got new 1) ..i.e replacing one by one ..how it's replacing one by one?! Im imagining it as capacitor .. lets assume the capacitor having 1 and once I get new one , I want to charge the capacitor to one but it's already one ...so how does the previous one replacing by the current new one?!!!

I'll try to paint a picture with words...A single register bit has no recollection of its previous values.  It only remembers the value most recently written to it.  When you write a 1 or 0, the previous values are gone forever.
Let's consider your capacitor-based register.  If the capacitor is already charged which represents Logic 1, then writing a 1 to it doesn't change the logic value.  It stored 1 before and it stores 1 now.
On the other hand, if you write 0 to it by connecting the capacitor to
Logic 0, that discharges the capacitor giving it a Logic 0 value.In practice, capacitors leak so at the electrical level the newly-written Logic 1 is stronger than the previous Logic 1.  The values stored in capacitor-based registers need to be refreshed periodically or they go away.
If you're interested in how registers and memory cells are implemented, take a look at Wikipedia's articles on Memory Cells and Flip-Flops.

@ryan: "...is that stupid question to ask like this? I mean ..am I stupid or not good to ask like those questions?!..."They say that there's no such thing as a stupid question (although some come close) -- and you are certainly not stupid to ask questions -- but I think we all have to take a deep breath and a step back.Suppose I decided I wanted to work as a theoretical physicist -- I could start bouncing around the internet reading bits and pieces -- and I could start firing off questions to members of physics communities like "What is the angular momentum of a spinning electron" and "Why do some quarks have color" -- but whatever answers I received wouldn't help me because I would be leaping in at too  high a level. It would be much better for me to look around for a basic physics class and start working my way up from the bottom.I think this is where you are on your quest to work in analog electronics -- you need to get a good foundation before delving further -- you've "got to learn to walk before you can run" as they say.

To follow up with what Max said,  you don't have to go to a school to learn this but you do need to approach this in a more systematic manner.One way is to read/view online tutorials.  https://www.electronics-tutorials.ws/ is a good site (I recommend you start with "DC Circuits") but you can find others by doing an internet search for "electronics tutorials".As Max mentioned in a recent article  https://www.eeweb.com/profile/max-maxfield/ar...best-way-to-begin-learning-analog-electronics, the ARRL handbook has good sections on electronics theory. The comments to that article mention some other good books.

I haven't looked at online courses recently but I suspect there are a number of good low-cost/free courses that you could learn from. The advantage of taking a course is that you have other people to learn with. The disadvantage is that you often have to complete it on a schedule.