[S2] Bitwise operation: high order bits

In this exercise, we will work with operation on bits. When we speak about the position of a bit, index 0 corresponds to lowest order bit, 1 to the second-lowest order bit, ...

In C source code, you can write a number in binary (base 2) by prefixing it via 0b., e.g. 0b11010 = 26.

This exercise will introduce some non-standard data types which guarantee that the variable has a fixed number of bits. Indeed, on some machines, a int could use 2, 4 or 8 bytes. Hence, if we want to perform bitwise operations, we have to know first on how many bits we are working.

For this, C introduces a new class of variable types :

  • int8_t (signed integer of 8 bits)
  • uint8_t (unsigned integer of 8 bits)
  • uint16_t (unsigned integer of 16 bits)

You can mix uint or int with bit-lengths 8, 16, 32 and 64). These types are defined in <stdint.h>


Write the body of a function get_3_leftmost_bits that returns the 3 high order bits of x.

For instance, if x=0b11011001..., the function should return a uint8_t containing 0b00000110.

If x=0b01100001..., the function should return a uint8_t containing 0b00000011.

#include <stdint.h>
uint8_t get_3_leftmost_bits(uint32_t x) {

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Author(s) Arthur van Stratum
Deadline No deadline
Submission limit No limitation
Category Tags S2

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