bit-manipulation

I am trying to construct an IP header. An IP header has the following fields: Version, IHL, DSCP etc. I would like to populate a Byte Array such that I can store the information in bytes. Where I get confused however is that the Version field is only 4 bits wide. IHL is also only 4 bits wide. How do I fit the values of both of those fields to be represented as a byte? Do I need to do bitshifting? E.g. Version = 4, IHL = 5. I would need to create a byte that would equal 0100 0101 = 45h or 69 decimal. (byte) (4 << 4) | 5 This shifts the value 4 to the left, then sets lower 4 bits to the value 5.

I got an algorithm that I need to solve. Unfortunately, I can't even find a clue about a solution. Please help me to understand how to solve this problem. The problem An imaginary pen input device sends hex data to the device to display a line with color. Examples Draw simple Green Line Set color to green, draw a line from (0,0) to (4000, 4000). Filled circle in this diagram indicates a pen down position, empty circle indicates pen up position. Input Data: F0A04000417F4000417FC040004000804001C05F205F20804000 Output: CLR; CO 0 255 0 255; MV (0, 0); PEN DOWN; MV (4000, 4000); PEN UP; I got the

Is char c2=i1<<8>>24; valid C syntax? (Where i1 is and unsigned integer) Additionally, will it yield the result of shifting i1 8 bits left and 24 bits right respectively? I am unpacking a char previously stored in i1 , along with three other chars. Code below: unsigned char b3 = 202; unsigned char b2 = 254; unsigned char b1 = 186; unsigned char b0 = 190; ... unsigned int i1=202; i1=i1<<8; i1=i1+254; i1=i1<<8; i1=i1+186; i1=i1<<8; i1=i1+190; ... char c1=i1>>24; char c2=i1<<8>>24; The syntax is fine (although hard to read) and it will be parsed as c2 = (i1 << 8) >> 24 . So it will left shift i1

If x > y, then this function will return 1, other wise return 0. so far i have int isitGreater(int x, int y) { return (((y+((~x)+1)) >> 31) & 1); but it's not working. Allowed ops: Legal ops: ! ~ & ^ | + << >> I'm sure I have the logic right, if X - Y and I get a negative number, that means y > x , so therefore the 32nd bit is a 1, so I shift that bit to the right 31 times and then "and" it with "1". edit: this does not work if x is negative, due to overflow. how can i fix this overflow problem without using conditional statements? Your code works fine for me. Please submit a valid question.

How to find the number of pairs whose difference is a given constant and their bitwise AND is zero? Basically, all (x,y) such that x-y = k; where k is a given constant and x&y = 0; An interesting problem. Let k n-1 ...k 1 k 0 be the the binary representation of k . Let l be the index of the smallest i such that k i =1 We can remark that a potential pair of solutions x and y must have all their bits i , i<l at zero. Otherwise the only way to have a difference x-y with its i th bit unset would be to have x i =y i =1 and x&y will not have its i th bit unset. Now we arrive at the first bit at one

This question already has an answer here: Why is the output -33 for this code snippet 3 answers I'm trying to get the value of an integer using Bitwise NOT, but i'm not getting what i expected. #include <stdio.h> int main(){ int i = 16; int j = ~i; printf("%d", j); return 0; } Isn't 16 supposed to be: 00000000000000000000000000010000 So ~16 is supposed to be: 11111111111111111111111111101111 Why i'm not getting what i expected and why the result is negative? This is what i'm trying to do: I have a number for exemple 27 which is: 00000000000000000000000000011011 And want to check every bit if

I'm trying to learn how to use bitwise operators on a given input but am not having much luck figuring out how to use them. Let's say I have this following octet: 11(01)0000 How would I extract the bits between the braces? You need to: create a suitable mask with ones only where are the bytes you need (you just need to write the number in binary and convert to e.g. hex to put it inside the C program). The parentheses in your 11(01)0000 are your indication to where to put the ones in your mask. Alternatively, create a mask made of as many ones as the chunk of bits you are interested in (in your

I'm in the unique situation where searching for "most significant bit" yields too many results and I can't find an answer that fits my needs! The question itself is pretty simple: "How do I find the most significant set bit in an unsigned long?" When I do my calculations the rightmost bit position is position '0'. I know that it involves masking the lowest bit, checking and then shifting left to once while incrementing my count, and then repeating with the 2nd lowest, etc. I've done this before but for whatever reason I can't do it now. Edit: By "most significant" I mean leftmost set bit,