Using a bitmask and if statement - c

Question

I am trying to allow multiple cases to run in a switch statement. I have a bitmask as follows:
#define SHOOT_ROCKET 2 << 16
#define MOVE_FORWARD 3 << 16
Later, I do
switch (int game_action)
and I have
case SHOOT_ROCKET:
result = fire_weapon(rl);
I don't want to 'break', because I want possibility of multiple actions. But I am returning a value called 'result'. I store this as a variable and return at the end. I can tell other case: statements are running though even when they shouldn't because result keeps getting changed and doesn't if I add break;
What is the best way to deal with this?
Update: I've been told to do if instead.
I changed my << bitmasks so they start at 1 now.
I am experiencing a weird bug
if (game_action->action & SHOOT_ROCKET)
{
game_action->data=5;
}
if (game_action->action & MOVE_FORWARD)
{
game_action->data=64;
}
I am not concerned about game_action being overwritten when I intend for multiple if's to evaluate to true
However: it seems MOVE_FORWARD is happening even if I only try and shoot a rocket!
game_action is a void pointer normally, so this is how it's setup in the function:
game_action = (game_action) *game_action_ptr;
I have verified the bitmask is correct with
printf("%d", game_action->action >> 16) which prints 2.
So why is '3' (the move forward) happening when I am only trying to shoot a rocket?

Answer 1

Please do update your question.
So why is '3' (the move forward) happening when I am only trying to shoot a rocket?
The first thing you want to look at is
#define SHOOT_ROCKET 2 << 16
#define MOVE_FORWARD 3 << 16
and think what that evaluates to. It will evaluate to the following binary numbers (Python is handy for this kind of stuff, 0b is just a prefix that means binary is following):
>>> bin(2 << 16)
'0b100000000000000000'
>>> bin(3 << 16)
'0b110000000000000000'
So you see that you use one bit twice in your #defines (Retired Ninja already pointed this out). This means that if game_action->action is set to anything where the bit 2 << 16 is 1, both of your ifs will evaluate to true, because both #defines have that bit set to 1.
to make them mutually exclusive, should i do 2, 4, 8, instead of 1,2,3,4?
If you want to easily keep track of which bits are used, you can either use powers of two (1,2,4,8,16, etc), e.g. #define MOVE_FORWARD 4 (I'm ignoring the << 16 you have, you can add that if you want), or you can shift a 1 by a variable number of bits, both result in the same binary numbers:
#define MOVE_LEFT 1 << 0
#define MOVE_RIGHT 1 << 1
#define MOVE_UP 1 << 3
#define MOVE_DOWN 1 << 4
There are legitimate cases where bitmasks need to have more than one bit set, for example for checking if any one of several bits are set:
#define MOVE_LEFT ... (as above)
#define SHOOT_ROCKET 1 << 5
#define SHOOT_GUN 1 << 6
//...
#define ANY_MOVEMENT 0xF
#define ANY_WEAPON_USE 0xF << 4
and then check:
if (action & ANY_MOVEMENT) { ... }
if (action & ANY_WEAPON_USE) { ... }

Answer 2

place parens '(' ')' around the value part (2<<15) kind of values so there is no errors introduced by the text replacement.
I.E. this:
'if( (&game_action->action & MOVE_FORWARD) == MOVE_FORWARD)'
becomes
'if( (&game_action->action & 2 << 16) == 2 << 16)'
Note the posted code is missing a left paren, which I added.
Where the '&' has a higher Precedence the '<<' so it (effectively) becomes
'if( ( (&game_action->action & 2) << 16) == 2 << 16)'
where the '&' is done to the 2 and not to the 2<<16