How does condition statement work with bit-wise operators?

Yes, you are right in the last part. Binary & and | are performed bit by bit. Since

1 & 1 == 1
1 & 0 == 0
0 & 1 == 0
0 & 0 == 0

we can see that:

8 & 1 == 1000 & 0001 == 0000

and

7 & 1 == 0111 & 0001 == 0001

Your test function does correctly compute whether a number is even or odd though, because a & 1 tests whether there is a 1 in the 1s place, which there only is for odd numbers.

Actually, in C, C++ and other major programming languages the & operator do AND operations in each bit for integral types. The nth bit in a bitwise AND is equal to 1 if and only if the nth bit of both operands are equal to 1.

For example:

8 & 1 =
1000 - 8
0001 - 1
----
0000 - 0

7 & 1 =
0111 - 7
0001 - 1
----
0001 - 1

7 & 5 =
0111 - 7
0101 - 5
----
0101 - 5

For this reason doing a bitwise AND between an even number and 1 will always be equal 0 because only odd numbers have their least significant bit equal to 1.

if(x) in C++ converts x to boolean. An integer is considered true iff it is nonzero.

Thus, all if(i & 1) is doing is checking to see if the least-significant bit is set in i. If it is set, i&1 will be nonzero; if it is not set, i&1 will be zero.

The least significant bit is set in an integer iff that integer is odd, so thus i&1 is nonzero iff i is odd.