I am using C for developing my program and I found out from an example code
unHiByte = unVal >> 8;
What does this mean?? If unVal = 250. What could be the value for unHiByte??
I am really confused with this..Please Help..
Thanks in advance..
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Nadeem_MK
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Ferin Jose
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[Shift Operators: >> and < – Frédéric Hamidi Oct 03 '13 at 07:34
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Read [What does this “>>=” operator mean in C?](http://stackoverflow.com/questions/17769948/what-does-this-operator-mean-inc/17769959#17769959) – Grijesh Chauhan Oct 03 '13 at 07:44
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1Why not simply try it out for yourself? – bitmask Oct 03 '13 at 07:47
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i think the OP is confused with the why i.e. the high level intention of using bit shifts. +1 for that possibility – fkl Oct 03 '13 at 09:15
4 Answers
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">>" in programming is a bitwise operation. The operation >> means shift right operation. So unVal >> 8 means shift right unVal by 8 bits. shifting the bits to the right can be interpreted as dividing the value by 2.
Hence, unHiByte = unval >> 8 means unHiByte = unVal/(2^8) (divide unVal by 2 eight times)
CodingBird
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Slight clarification >> 1 means dividing by 2 once. First part of answer makes it look like >> 8 means dividing by 2 once. +1 – fkl Oct 03 '13 at 07:41
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@fayyazkl sorry about that, maybe its just my bad way of explaining. haha – CodingBird Oct 03 '13 at 08:08
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@LekshmiGopikrishnan no prob! :) accept the answer if its what you want please. :) – CodingBird Oct 03 '13 at 08:10
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@CodingBird By no means that is a problem. We all just collaborate to help each other. The concept is right. I just suggested a wording improvement with a possible confusion. – fkl Oct 03 '13 at 08:17
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Without going into the shift operator itself (since that is answered already), here the assumption is that unVal is a two byte variable with a high byte (the upper 8 bits) and a low byte (the lower 8 bits). The intent is to obtain the value produced by ONLY the upper 8 bits and discarding the lower bits.
The shift operator though should easily be learned via any book / tutorial and perhaps was the reason some one down voted the question.
fkl
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OK so unHiByte = unVal << 8; means retrieving only the lower 8 bits from unVal?? Please make that too clear for me. – Ferin Jose Oct 03 '13 at 08:50
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Not exactly. The concept is right though. But the problem is when you shift some thing left, it appends 8 zeros in the lower byte and promote the lower byte to higher one. To obtain lower bits only, & it will a mask with only right most 8 bits as 1 i.e. unVal & 255. – fkl Oct 03 '13 at 09:14
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The >> is a bitwise right shift.
It operates on bits. With unHiByte = unVal >> 8; When unVal=250.
Right shift means to shift the bits to the right. So when you shift 1111 1010, 8 digits to right you get 0000 0000.
Note: You can easily determine the right shift operation result by dividing the number to the left of >> by 2^(number to right of >>)
So, 250/28= 0
Suvarna Pattayil
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Not a single person over the internet has been able to describe this simple thing in easy way, All just throw definition of it not application that why is it used for. Here is the understanding.
For example: if you have a hex 0x2A63 and you want to take 2A or you want to take 63 out from it, then you will do this.
For example FYI if we covert 2A63 to binary which is: 0010101001100011. (that is 16 bits) (first 8 bits are 2A and second bits are 63) The issue is that binary always starts from right. So we have to push the first 8 bit which (2A) to the right side to be able to get it.
Line 1: uint16_t hex = 0x2A63;
Line 2: uint8_t part2A = (uint8_t)(hex >> 8) // Pushed the first // eight bits (2A) to right and (63) is gone out of the way. Now we have 0000000000101010
// Now Line 2 returns for us 0x2A which the last 8 bits (2A).
Line 3: // To get 63 we will do simply:
uint8_t part63 = (uint8_t)hex; // As by default the 63 is in the right most side in the binary.
It is that simple.
