How does it work "checking bit flag"

Question

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?

Answer 1

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.

Answer 2

An int can be converted to a bool. If its value is zero, the converted value ist false, otherwise it is true. Check it out yorself:

bool f = 0;
bool t = 5;

Prvalues of integral, floating-point, unscoped enumeration, pointer, and pointer-to-member types can be converted to prvalues of type bool.

The value zero (for integral, floating-point, and unscoped enumeration) and the null pointer and the null pointer-to-member values become false. All other values become true.

See: cppreference

Answer 3

The & operator performs a bitwise boolean AND operation. That means that a bit is set in the final output if, and only if, it is set in both inputs, so for example

  00011100
& 01010101
  --------
= 00010100

  11010101
& 01010101
  --------
= 01010101

So, if you use a mask that has only a single bit set it will return 0 if that bit is not set, and non-zero if it is set, e.g.

  11110111
& 00001000
  --------
= 00000000

  00001111
& 00001000
  --------
= 00001000

Then, since C++ will convert treat any non-zero value as true when converting to a bool and zero as false the function works.

Answer 4

Since this is also tagged C, C99 has this to say:

6.3.1.2 Boolean type

1 When any scalar value is converted to _Bool, the result is 0 if the value compares equal to 0; otherwise, the result is 1.