与C ++中的枚举有关的问题。 这是一个已知问题吗？

Question

I have found a problem with enum in C++. I wonder if that is a known issue.

#include <iostream>

/* run this program using the console pauser or add your own getch,
   system("pause") or input loop */

using namespace std;

enum color { red = 8, green = 7, blue };

int main(int argc, char** argv)
{
    color r = red, g = green, b = blue; 

    cout << r << " " << g<< " " << b << " " <<endl;

    switch (b) {
        case red:
            cout << "a bad thing happened" << endl;
            break;

    }

    return 0;
}

running the program you get:

8 7 8 a bad thing happened

Answer 1

Quoting from cppreference.com

If the first enumerator does not have an initializer, the associated value is zero. For any other enumerator whose definition does not have an initializer, the associated value is the value of the previous enumerator plus one.

So basically blue is green +1 => 8 which is equal to red (8)

Answer 2

Yes, b is 8, same as r . In enums, if you don't specify the exact value for an enumerator, it will be one more than the previous one, 7+1=8 in your case.