为什么直接比较两个枚举时会收到错误？

C规范(6.7.2.2)的枚举特定部分规定：

The identifiers in an enumerator list are declared as constants that have type int and
may appear wherever such are permitted.¹²⁷⁾ An enumerator with = defines its
enumeration constant as the value of the constant expression. If the first enumerator has
no =,the value of its enumeration constant is 0.

所以我应该能够使用与int常量相同的枚举成员.在这个小样本程序中：

enum first {
  a,b
};

enum second {
 c,d
};

int main(){
    enum first myf = a;
    enum second mys = c;

    if(myf == mys)
        printf("same valuen"); 
    return 0;
}

当编译gcc -Wall -Werror时,我收到消息：

error: comparison between ‘enum first’ and ‘enum second’ [-Werror=enum-compare]

我知道,如果我将myf和mysql的类型转换为int,那么编译器将会很开心,就像我可以用myf和mys的值来设置几个int,并进行比较;但为什么我要做这些任何一个来摆脱警告呢？为什么首先存在这个警告？这样做,我看不到会有一些危险.

注意：我已经阅读了关于这个枚举比较标志的gcc文档,但它并没有说什么：

-Wenum-compare
Warn about a comparison between values of different enumerated types. In C++ enumeral mismatches in conditional expressions are also diagnosed and the warning is enabled by default. In C this warning is enabled by -Wall.