Let n be the position of the leftmost 1 bit if x. … So when we say x 1, we are saying "shift the bits in the variable x left by 1 place". The high-order empty bit positions are set based on the type of the left-hand operand as follows: If the left-hand operand is of type int or long, the right-shift operator performs an arithmetic shift: the value of the most significant bit (the sign bit) of the left-hand operand is propagated to the high-order empty bit positions. Right shift is equivalent to dividing the bit pattern with 2k ( if we are shifting k bits ). New bits shifted in from the right side receive the value 0. The << (left shift) in C or C++ takes two numbers, left shifts the bits of the first operand, the second operand decides the number of places to shift. 0011 << 1 is 0110 0011 << 2 is 1100 0011 << 3 is 1000. The bitwise shift operators are used to move/shift the bit patterns either to the left or right side. The result may be expressed by the formula 1U << (lg(v - 1) + 1). Bits that are shifted off the end of the binary number … Left and right are two shift operators provided by 'C' which are represented as follows: Operand << n (Left Shift) Operand >> n (Right Shift) When 1 is added to ~x, all positions below n become 0 and the 0 at position n becomes 1. These operators cause the bits in the left operand to be shifted left or right by the number of positions specified by the right operand. If we wanted to shift it to the left 2 places, we'd end up with 00100000; everything is moved to the left two places, and zeros are added as padding. If x is a power of two, its lone 1 bit is in position n. This means ~x has a 0 in position n and 1s everywhere else. This results in double the previous number. variable << number_of_bits variable >> number_of_bits … int mult_by_pow_2(int number, int power) { return number<> 1 = 2 6 >> 1 = 3 5 >> 1 = 2 16 >> 4 = 1 A << 2 = 240 i.e., 1111 0000 >> Binary Right Shift Operator. This is done by using the left shift operator and shifting the bits left by 1. AND - Value of c is 20 OR - Value of c is 21 Exclusive-OR - Value of c is 1 Bitwise shift operators. 2. The left operands value is moved right by the number of bits specified by the right operand. The result of XOR is 1 if the two bits are different. More on bitwise math may be found here. In 12 operations, this code computes the next highest power of 2 for a 32-bit integer. A >> 2 = 15 i.e., 0000 1111 In other words, the carry propagates all the way to position n. So what happens is … The left-shift and right-shift operators are equivalent to multiplication and division by 2 respectively. The left operands value is moved left by the number of bits specified by the right operand. This is the number 32 -- in fact, left shifting is the equivalent of multiplying by a power of two. Note that in the third case, we shifted a bit off the end of the number! A number can be multiplied by 2 using bitwise operators. Live Demo Binary Left Shift Operator. The ^ (bitwise XOR) in C or C++ takes two numbers as operands and does XOR on every bit of two numbers. Checking if given 32 bit integer is power of 2: A program that demonstrates multiplication of a number by 2 using bitwise operators is given as follows. There are two bit shift operators in C++: the left shift operator << and the right shift operator >>. In the divide version, you can see the idivl %r8d - but just above that is a cltd (convert long to double) and some additional logic around the spill and reload. Note that in the edge case where v is 0, it returns 0, which isn't a power of 2; you might append the expression v += (v == 0) to remedy this if it matters. In the bit shift version the key instruction is shll $2, %eax which is a shift left logical - there's the divide, and everything else is just moving values around. 1 << 4 = 16 = 2 4 … 1 << n = 2 n. Right Shift ( >> ): Right shift operator is a binary operator which shift the some number of bits, in the given bit pattern, to the right and append 1 at the end. Example. This can be replaced with a left shift and an xor if the shift is wider than the number of bits you want to add, easy example is (i<<1)^1, which adds one to a doubled value. Syntax. This does not of course apply to a right shift (power of two divide) because only a left (little endian) shift fills the gap with zeros.