What does 0b1 mean in C++?

Question

I came across a part of code that I cannot understand.

for (unsigned int i = (x & 0b1); i < x; i+= 2)
    {
        // body
    }

Here, x is from 0 to 5.

What is meant by 0b1? and what would be the answers for eg: (0 & 0b1), (4 & 0b1) etc?

Answer 1

0b... is a binary number, just like 0x... is hex and 0... is octal.

Thus 0b1 is same as 1.

1b0 is illegal, the first digit in those must always be 0.

Answer 2

As previous answers said, it is the binary representation of the integer number 1, but they don't seem to have fully answered your question. This has a lot of layers so I'll briefly explain each.

In this context, the ampersand is working as a bitwise AND operator. i & 0b1 is (sometimes) a faster way of checking if an integer is even as opposed to i % 2 == 0.

Say you have int x = 5 and you'd like to check if it's even using bitwise AND.

In binary, 5 would be represented as 0101. That final 1 actually represents the number 1, and in binary integers it's only present in odd numbers. Let's apply the bitwise AND operator to 5 and 1;

 0101
 0001
&----
 0001

The operator is checking each column, and if both rows are 1, that column of the result will be 1 – otherwise, it will be 0. So, the result (converted back to base10) is 1. Now let's try with an even number. 4 = 0100.

 0100
 0001
&----
 0000

The result is now equal to 0. These rules apply to every single integer no matter its size.

The higher-level layer here is that in C, there is no boolean datatype, so booleans are represented as integers of either 0 (false) or any other value (true). This allows for some tricky shorthand, so the conditional if(x & 0b1) will only run if x is odd, because odd & 0b1 will always equal 1 (true), but even & 0b1 will always equal 0 (false).