Specifying size of enum type in C

问题:

Already read through this related question, but was looking for something a little more specific.

• Is there a way to tell your compiler specifically how wide you want your enum to be?
• If so, how do you do it? I know how to specify it in C#; is it similarly done in C?
• Would it even be worth doing? When the enum value is passed to a function, will it be passed as an int-sized value regardless?

回答1:

Is there a way to tell your compiler specifically how wide you want your enum to be?

In general case no. Not in standard C.

Would it even be worth doing?

It depends on the context. If you are talking about passing parameters to functions, then no, it is not worth doing (see below). If it is about saving memory when building aggregates from enum types, then it might be worth doing. However, in C you can simply use a suitably-sized integer type instead of enum type in aggregates. In C (as opposed to C++) enum types and integer types are almost always interchangeable.

When the enum value is passed to a function, will it be passed as an int-sized value regardless?

Many (most) compilers these days pass all parameters as values of natural word size for the given hardware platform. For example, on a 64-bit platform many compilers will pass all parameters as 64-bit values, regardless of their actual size, even if type int has 32 bits in it on that platform (so, it is not generally passed as "int-sized" value on such a platform). For this reason, it makes no sense to try to optimize enum sizes for parameter passing purposes.

回答2:

I believe there is a flag if you are using GCC.

-fshort-enums

回答3:

You can force it to be at least a certain size by defining an appropriate value. For example, if you want your enum to be stored as the same size as an int, even though all the values would fit in a char, you can do something like this:

typedef enum {
    firstValue = 1,
    secondValue = 2,

    Internal_ForceMyEnumIntSize = MAX_INT
} MyEnum;

Note, however, that the behavior can be dependent on the implementation.

As you note, passing such a value to a function will cause it to be expanded to an int anyway, but if you are using your type in an array or a struct, then the size will matter. If you really care about element sizes, you should really use types like int8_t, int32_t, etc.

回答4:

There is also another way if the enum is part of a structure:

struct something {
     :0;
   enum whatever field:CHAR_BIT;
     :0;
};

The :0; can be omitted if the enum field is surrounded by normal fields. If there's another bitfield before, the :0 will force byte alignement to the next byte for the field following it.

回答5:

It depends on the values assigned for the enums.

Ex: If the value greater than 2^32-1 is stored, the size allocated for the overall enum will change to the next size.

Store 0xFFFFFFFFFFFF value to a enum variable, it will give warning if tried to compile in a 32 bit environment (round off warning) Where as in a 64 bit compilation, it will be successful and the size allocated will be 8 bytes.

回答6:

Another way is to cast the enum inside a union as follow:

union char_size {
  char a;
  enum {
    a = 1,
    b = 2,
  } val;
};

Doing so will force the compiler to fit the enum inside a char.