Question

for (unsigned int i = 1; i <= 100; i++) {
    if (i & 0x00000001) {
        std::cout << i<<",";
    }
}

why does (and how): if( i & 0x00000001 ) figure out the odd number?

Answer 1

0x00000001 is 1 in binary, although it’s written in hexadecimal (base-16) notation. That’s the 0x part.

& is the bit-wise ‘AND’ operator, which is used to do binary digit (bit) manipulations.

i & 1 converts all of the binary digits of i to zero, except for the last one.

It’s straightforward to convert the resulting 1-bit number to a boolean, for evaluation by the if statement.

The following chart shows the last 16 binary digits of i, and what happens to them.

i:   i in binary:        i & 1 in binary:    convert to boolean
---- ------------------- ------------------- ---------------------
1    0000000000000001    0000000000000001    true
2    0000000000000010    0000000000000000    false
3    0000000000000011    0000000000000001    true
4    0000000000000100    0000000000000000    false
5    0000000000000101    0000000000000001    true
6    0000000000000110    0000000000000000    false
7    0000000000000111    0000000000000001    true
8    0000000000001000    0000000000000000    false
...  ...                 ...                 ...
99   0000000001100011    0000000000000001    true
100  0000000001100100    0000000000000000    false