cannot initialize a variable of type 'designFlags' with an rvalue of type 'int'

Question

i am trying to create an enum flag that would see if a certain thing was bold and italic, but i am getting compilation error: error: cannot initialize a variable of type 'designFlags' with an rvalue of type 'int' designFlags myDesign = BOLD | UNDERLINE;

#include <iostream>
using namespace std;

enum designFlags {
    BOLD = 1,
    ITALICS = 2,
    UNDERLINE = 4
};

int main() 
{
    designFlags myDesign = BOLD | UNDERLINE; 

        //    00000001
        //  | 00000100
        //  ___________
        //    00000101

    cout << myDesign;

    return 0;
}

Answer 1

The C++ standard prevents you from performing an implicit cast from int to enum (Note: implicit conversion from enum to int is allowed).

However, you can perform an static cast instead:

designFlags myDesign = static_cast<designFlags>(BOLD | UNDERLINE);
cout << myDesign; // prints 5

Have in mind that it is most like that the resulting value is not one of the values specifies in designFlags. This is usually a common practice, but you have to rely to bit-wise operations to perform comparisons:

For example:

// wrong
if (myDesign == BOLD)

// ok (most of the time)
if (myDesign & BOLD)

// always ok
if ((myDesign & BOLD) == BOLD)

Update: Why is myDesign & BOLD ok most of the time?

When you use flags, the values must be powers of two: 1, 2, 4, 8, etc. Therefore, every possible value has only one bit set, and this bit is different for each one.

Then, when you do myDesign & BOLD, you are testing for one specific bit, and just that. If the bit is set, the result will be non-zero (bitwise AND). If the bit it not set, the result will be zero.

When it can go wrong? Well, if you use flags that are not a power of two, each individual value will have more than one bit set, making it difficult to perform these checks.

For instance, suppose you have this definition:

enum designFlags {
    BOLD = 2,
    ITALICS = 4,
    UNDERLINE = 6
};

And then you set your design to UNDERLINE:

myDesign = UNDERLINE

What is the result of the following tests?

myDesign & BOLD
myDesign & ITALICS

It will be non-zero for both of them, surely an undesired behavior.

Of course you can do the following:

if ((myDesign & BOLD) == BOLD) {
    ...

but i rather use flags with only one bit set.