What is (x & 1) and (x >>= 1)?

Question

I am trying to do assignment: "Find the number of bits in an unsigned integer data type without using the sizeof() function."

And my design is to convert the integer to bits and then to count them. For ex: 10 is 1010 and 5 is 101

Converting integer to a bit representation shows something like this:

do
{ 
    Vec.push_back( x & 1 ) 
} 
while ( x >>= 1 );

I don't want to just copy paste stuff. When I use F-10 I see what (x & 1) is doing but I don't know it is name or how it does its job(compare something?). Also I know >= which "greater than or equal" but what is x >>= 1?

Note: The marked duplicate is a JavaScript and not C++

Solution 1

These are Bitwise Operators (reference).

x & 1 produces a value that is either 1 or 0, depending on the least significant bit of x: if the last bit is 1, the result of x & 1 is 1; otherwise, it is 0. This is a bitwise AND operation.

x >>= 1 means "set x to itself shifted by one bit to the right". The expression evaluates to the new value of x after the shift.

Note: The value of the most significant bit after the shift is zero for values of unsigned type. For values of signed type the most significant bit is copied from the sign bit of the value prior to shifting as part of sign extension, so the loop will never finish if x is a signed type, and the initial value is negative.

Solution 2

x & 1 is equivalent to x % 2.

x >> 1 is equivalent to x / 2

So, these things are basically the result and remainder of divide by two.

Solution 3

In addition to the answer of "dasblinkenlight" I think an example could help. I will only use 8 bits for a better understanding.

x & 1 produces a value that is either 1 or 0, depending on the least significant bit of x: if the last bit is 1, the result of x & 1 is 1; otherwise, it is 0. This is a bitwise AND operation.

This is because 1 will be represented in bits as 00000001. Only the last bit is set to 1. Let's assume x is 185 which will be represented in bits as 10111001. If you apply a bitwise AND operation on x with 1 this will be the result:

00000001
10111001
--------
00000001

The first seven bits of the operation result will be 0 after the operation and will carry no information in this case (see Logical AND operation). Because whatever the first seven bits of the operand x were before, after the operation they will be 0. But the last bit of the operand 1 is 1 and it will reveal if the last bit of operand x was 0 or 1. So in this example the result of the bitwise AND operation will be 1 because our last bit of x is 1. If the last bit would have been 0, then the result would have been also 0, indicating that the last bit of operand x is 0:

00000001
10111000
--------
00000000

x >>= 1 means "set x to itself shifted by one bit to the right". The expression evaluates to the new value of x after the shift

Let's pick the example from above. For x >>= 1 this would be:

10111001
--------
01011100

And for left shift x <<= 1 it would be:

10111001
--------
01110010

Please pay attention to the note of user "dasblinkenlight" in regard to shifts.

Solution 4

It is similar to x = (x >> 1).

(operand1)(operator)=(operand2)  implies(=>)  (operand1)=(operand1)(operator)(operand2) 

It shifts the binary value of x by one to the right.

E.g.

int x=3;    // binary form (011) 
x = x >> 1; // zero shifted in from the left, 1 shifted out to the right:
            // x=1, binary form (001)

Solution 5

(n & 1) will check if 'n' is odd or even, this is similar to (n%2).

1. In case 'n' is odd (n & 1) will return true/1;

2. Else it will return back false/0;

---

The '>>' in (n>>=1) is a bitwise operator called "Right shift", this operator will modify the value of 'n', the formula is:

(n >>= m) => (n = n>>m) => (n = n/2^m)

Go through GeeksforGeek's article on "Bitwise Operator's", recommended!

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