Cout with &

Question

Refer below code:

#include <iostream>
using namespace std;

int main() {

for(int i=0;i<10;++i)
{
if(i&1)
{
cout << i << endl;
}
}
return 0;
}

Why output of same is 1 3 5 7 9 ? could not get statement of if....

Answer 1

As Ipang stated, the if statement condition is a simple form of bit hacking to extract odd numbers and it's equivalent to if (i % 2 != 0).

The for loop cycles 10 times (0-9). Let's convert each number to its binary form, since I wanna explain it in more detail of course and more than that it's a BITwise operation!

4 bits is enough for representing the whole range.

0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001

Now let's say, we wanna evaluate number 7. What we are gonna do is, nothing but ANDing binary form of number 7 with 1 (I suppose you know how to do that).

0111
0001 &
________
0001

Yay! you got 1 which means true which means it's an odd number. what if it is an even one? Let's say 4.

0100
0001 &
________
0000

Whoops! you got 0 which means it's an even number.

Answer 2

It filters out the even numbers so you get the odd ones only.

Answer 3

C++ Soldier (Babak) just gave an excellent explanation in detail, I was only giving a short footage of what it does.

Big Thanks : )

Answer 4

C++ Soldier (Babak) No mate! not at all, I never thought that way, as a matter of fact, was actually grateful you came up with that, to explain how it works under the hood, how and why it worked out the way it did, I value people giving explanatory answers, no issues whatsoever mate : )

Cheers Babak ; )

Answer 5

Ketan Lalcheta
Compilers use complex grammars to identify what the heck is going on in a program in different contexts. For example, & operator in [ int *y = &x; ] is considered the address (reference) operator but in [ bool b = x & y; ], bitwise AND. You see, it has some sort of analyzer and predefined patterns by which it can confidently recognizes that the operator & has 2 operand or one (Of course, our case here is easy to recognize).

Answer 6

Oh come on Ipang !
I didn't do that to render your answer less worthy. You just gave me the motivation to make it a bit lengthy! ; - ]

Answer 7

Ipang
Kind and reasonable as always. SL would've been get dull without people like you, dear. Now let me mess with this post a bit! I guess OP don't mind after all!

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Answer 8

thanks a lot for detailed explanation Ipang....any idea why compiler considered it as bitwise operation? does it due to the & operator added into if condition??