Flags, enum (C) | 易学教程

问题

I'm not very used to programming with flags, but I think I just found a situation where they'd be useful:

I've got a couple of objects that register themselves as listeners to certain events. What events they register for is dependent on a variable that is sent to them when they are constructed. I think a nice way to do this would be to send bitwise OR connected variables, like such: TAKES_DAMAGE | GRABBABLE | LIQUID, etc. Then, in the constructor, the object can check what flags are set and register it self as listener for the ones that are.

But this is where I get confused. Preferably, the flags would be in an enum. But that is also a problem. If we have got these flags:

enum
{
    TAKES_DAMAGE,/* (0) */
    GRABBABLE, /* (1) */
    LIQUID, /* (2) */
    SOME_OTHER /* (3) */
};

Then sending the flag SOME_OTHER (3) will be the same as sending GRABBABLE | LIQUID, will it not?

How exactly do you deal with this stuff?

回答1:

Your enumeration needs to be powers of two :

enum
{
    TAKES_DAMAGE = 1,
    GRABBABLE = 2,
    LIQUID = 4,
    SOME_OTHER = 8
};

Or in a more readable fashion :

enum
{
    TAKES_DAMAGE = 1 << 0,
    GRABBABLE = 1 << 1,
    LIQUID = 1 << 2,
    SOME_OTHER = 1 << 3
};

Why ? Because you want to be able to combine flags with no overlapping, and also be able to do this:

if(myVar & GRABBABLE)
{
    // grabbable code
}

... Which works if the enumeration values look like this :

 TAKES_DAMAGE: 00000001
 GRABBABLE:    00000010
 LIQUID:       00000100
 SOME_OTHER:   00001000

So, say you've set myVar to GRABBABLE | TAKES_DAMAGE, here's how it works when you need to check for the GRABBABLE flag:

 myVar:     00000011
 GRABBABLE: 00000010 [AND]
 -------------------
            00000010 // non-zero => converts to true

If you'd set myVar to LIQUID | SOME_OTHER, the operation would have resulted in :

 myVar:     00001100
 GRABBABLE: 00000010 [AND]
 -------------------
            00000000 // zero => converts to false

回答2:

another way of storing flags is to not bother with the underlying type at all. when using an enum, the enum values are stored by default into an unsigned int, which is 32 bits on a common computer. this gives you with only 32 possible flags: while certainly much, there are some cases where it is not sufficient.

now you can define your flag set this way:

typedef struct
{
    int takes_damage : 1;
    int grabbable    : 1;
    int liquid       : 1;
    int some_other   : 1;
} flags;

if you never encountered this, the ': 1' part tells the compiler to only use 1 bit to store this struct member.

now you can define a variable to hold the flags, and work with those flags:

flags myflags = {1,0,0,1}; // defines a variable holding a set of flags, with an initial value of takes_damage & some_other

myflags.liquid = 1; // change the flags to include the liquid

if ( myflags.takes_damage ) // test for one flag
    apply_damage();
if ( myflags.liquid && myflags.some_other ) // test for multiple flags
    show_strange_behavior();

this method allows you to define any number of flags, without limitation, and you can extend your flag set at any time without fearing an overflow. the drawback is that testing a subset of the flags is more cumbersome and necessitate more code.

回答3:

Yes. Instead, make your enum members powers of 2:

enum
{
    TAKES_DAMAGE = (1 << 0),
    GRABBABLE = (1 << 1),
    LIQUID = (1 << 2),
    SOME_OTHER = (1 << 3)
};

回答4:

You should make the flags only powers of two, i.e. each is a bit in whatever data type you're storing this in, and nothing overlaps when you bitwise OR.

回答5:

Can't you just set the values in the enum?

enum {
 TAKES_DAMAGE = 1,
 GRABBABLE    = 2,
 LIQUID       = 4
}

Afterwards, just perfom bit-wise OR on them.

回答6:

you need

enum
{
    TAKES_DAMAGE = 1,
    GRABBABLE = 2,
    LIQUID = 4,
    SOME_OTHER = 8
};

来源：https://stackoverflow.com/questions/1631266/flags-enum-c

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