How does condition statement work with bit-wise operators?
bitwise-operators, c++
Solution
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.
Problem
I tried to understand how if condition work with bitwise operators. A way to check if a number is even or odd can be done by: ``` #include <iostream> #include <string> using namespace std; string test() { int i = 8; //a number if(i & 1) return "odd"; else return "even"; } int main () { cout << test(); return 0; } ``` The Part I don't understand is how the if condition work. In this case if i = 8 then the in If statement it is doing `1000 & 1` which should gives back 1000 which equal 8. If i = 7, then in if statement it should be doing `111 & 1` which gives back 111 which equal 7 Why is it the case that if(8) will return "even" and if(7) return "odd"? I guess I want to understand what the if statement is checking to be True and what to be False when dealing with bit-wise operators. Just A thought when I wrote this question down is it because it's actually doing ``` for 8: 1000 & 0001 which gives 0 for 7: 0111 & 0001 which gives 1? ```