How does it work "checking bit flag"

bit-manipulation, bit-shift, c++

Solution

The C++ draft standard section `4.12` Boolean conversions says (emphasis mine):

A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true. For direct-initialization (8.5), a prvalue of type std::nullptr_t can be converted to a prvalue of type bool; the resulting value is false.

So any non-zero value will be converted to `true`.

You expressed some doubts as to the result of `1 << 4` we have a quick way to check this using std::bitset:

#include <bitset>
#include <iostream>

int main() 
{
    std::bitset<8> b1(1<<4);

    std::cout << b1 << std::endl ;

    return 0;
}

and we can see the result is indeed `00010000` since bitwise and only set a bit to `1` if the bit in both operands are `1` then the result of `flag & (1 << 4)` will only be non-zero if bit `5` of `flag` is also `1`.

Problem

I've read other questions about this, but they don't answer my question. In the following code I understand that it checks if that bit is set, but my question is why? ``` bool test(uint8_t& flag) { return flag & (1 << 4); } ``` I don't get why it returns a `bool` and it works, flag is `uint8_t` and `1 << 4` should be something like this `00010000` (I think). Why does that code return the value of the desired single bit and not the rightmost or something like that?

Original source