In the Linux kernel tree, the file /include/linux/sched.h contains task_struct which, among other data, defines a variable volatile long state.
According to this page, the number stored in state represents one of five states:
#define TASK_RUNNING 0
#define TASK_INTERRUPTIBLE 1
#define TASK_UNINTERRUPTIBLE 2
#define TASK_ZOMBIE 4
#define TASK_STOPPED 8
My question is as follows:
Why is such a large datatype used to store the state? Wouldn't a single byte be perfectly sufficient to store the states above? Surely, long wasn't chosen in anticipation of thousands/millions of possible states?
I'm aware that the definition of long will vary across the different architectures, but I thought char would have sufficed.
You can easily see these values are power-of-2. This is the standard way to handle /flags/. You can logically OR ('|') them to create cobined states. And then you can easily test them with AND ('&').
long on a 32-bit platform will only be 32-bit.
So, with long on a 32-bit platform you only have 32 distinct flags, as opposed to millions.
Finally, I strongly suggest inspecting actual sources and preferably from this decade, see:
#define TASK_RUNNING 0
#define TASK_INTERRUPTIBLE 1
#define TASK_UNINTERRUPTIBLE 2
#define __TASK_STOPPED 4
#define __TASK_TRACED 8
/* in tsk->exit_state */
#define EXIT_DEAD 16
#define EXIT_ZOMBIE 32
#define EXIT_TRACE (EXIT_ZOMBIE | EXIT_DEAD)
/* in tsk->state again */
#define TASK_DEAD 64
#define TASK_WAKEKILL 128
#define TASK_WAKING 256
#define TASK_PARKED 512
#define TASK_NOLOAD 1024
#define TASK_STATE_MAX 2048
And flag usage below, e.g.
#define TASK_KILLABLE (TASK_WAKEKILL | TASK_UNINTERRUPTIBLE)
So, the question is why would this be long instead of an int. That I don't know, but it is unlikely there is any particularly good reason.
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