occurrences
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Write the function int count101(int n) that counts the number of occurrences of the bit pattern 101 in n.
It should return the final count. Variable n is 32 bits. Example: Suppose n is just 8 bits, say n=11010101, then the function would return 3.
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- In C, write a function int setbit(int n, int i) to set the i^ᵗʰ bit of n if i^ᵗʰ bit is 0.Write a function int bitset(int x, int n) to set nth bit of x to 1, if its (n-1)th bit is 1. if n <1 or n >15, the function will return -1 to thecalling functionWrite a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.
- Write a function timestable(n), which prints a multiplication table of size n. For example, timestable(5) would print: 1 2 3 4 5 2 3 4 5 4 6 8 10 6 9 12 158 12 16 2010 15 20 25Use the function in a program where you ask the user for n and you print the corresponding table.Write a function that makes the base conversion. It takes two integers n and b as input (n >= 0 andb > 1, both in base 10) and expresses n in base b. For example, the decimal expansion of 345 inbase 10 is 345 = 3 × 102 + 4 × 101 + 5 × 100.Sample input: n = 12, b =2. Sample output: 1100Explanation: (12)10 = (1100)2Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required.Number of loops isequal to the number of 1s in the binary representation."""def count_ones_recur(n):.
- Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required..Write a C-function with two arguments (n and r) that has prototype: char clearbit(char k, char bits) The function clears (sets to 0) the bit number k (in the range of 0 to 7) in bits and returns the resulting value. For example, if k is 0x02 and bits is 0x07, the function would return bits with its k’th bit cleared, resulting in 0x03. It must not change other bits in bits. Hint: You may use any number of C-statements, but this task can be accomplished in as few as one!c language Write a function which takes an integer and returns 1 if there are an odd number of ‘1’ bits 0 if there are an even number of ‘1’ bits Example, of x is 1010 in binary, it should return 0 because there are two 1's. 1110 should return 1 because there are three 1's
- Consider the following code which includes a parity check for the third digit: C=(000,011,101,110). You get the third digit by adding the first two and then using the remainder on division by 2. Can this code detect and count single errors? how can this be implemented?implement anyEvenBit(x) Return 1 if any even bit in x is set to 1 you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * anyEvenBit - return 1 if any even-numbered bit in word set to 1* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1* Legal ops: ! ~ & ^ | + << >>* Max ops: 12*/int anyEvenBit(int x) {return 2;}Implement the function (in C or C++) with the following prototype: /** Implement a function which rotates a word left by n-bits, and returns that rotated value. Assume 0 <= n < w Examples: when x = 0x12345678 and w = 32: n=4 -> 0x23456781 n=20 -> 0x67812345 */ unsigned rotate_left(unsigned x, int n); The function should follow the bit-level integer coding rules above. Be careful of the case n = 0.