C++ typecasting with bitwise assignement operation - c++

Question

I haven't found a way to typecast where bitwise assignment operation (e.g. x |= y &z;) is involved.
Example:
#include <stdio.h>
typedef enum type1{
AA = 0,
BB = 1
} type1_e;
int main()
{
type1_e x,y;
y = type1_e(y | x); //working
y |= (type1_e)(y|x); //ERROR: as.cc:15: error: invalid conversion from ‘int’ to ‘type1_e’
}

Answer 1

operator | yields an int result
type1_e(y | x)
y | x is an int. You're explicitly casting that int to type1_e.
y |= (type1_e)(y|x);
is equivalent to
y = y | type1_e(y | x);
You are using operator | which yields an int result, then trying to assign that to y which is type1_e. You can't do that without casting.
To overcome that, you could do this:
y = type1_e(y | type1_e(y | x));
which is the same as:
y = type1_e(y | y | x);
which is the same as:
y = type1_e(y | x);
or:
y = static_cast<type1_e>(y | x);

Answer 2

You can write an overload for bitwise OR (|) that works for your enum.
You will have to use static_cast as well for proper conversion.
#include<iostream>
using namespace std;
typedef enum type1{
AA = 0,
BB = 1
} type1_e;
type1_e operator |=(type1_e& x, type1_e y)
{
return x=static_cast<type1_e>(static_cast<int>(x) | static_cast<int>(y));
}
int main()
{
type1_e x = AA, y = BB;
y = type1_e(y | x); //working
std::cout << y << '\n';
x = AA, y = AA;
y |= static_cast<type1_e>(y|x); //also works
std::cout << y << '\n';
}
See demo.