Question

Write a function in XV6 that wakes up a single process. Use the function below as a reference.

// Wake up all processes sleeping on chan.
// The ptable lock must be held.
static void wakeup1 (void* chan) {
struct proc✶ p;
for (p = ptable. proc; p < &ptable.proc [NPROC]; p++)
if(p->state == SLEEPING && p->chan == chan)
p->state = RUNNABLE;
}
void wakeup_single (void* chan) {
// to implement
}

Answer 1

To implement the 'wakeup_single' function in XV6, you can use the 'wakeup1' function as a reference. It loops through the process table, checks if each process is in the 'SLEEPING' state and if its channel matches the given 'chan', and sets its state to 'RUNNABLE'.

Step-by-step explanation:

To implement the 'wakeup_single' function in XV6, you can use the 'wakeup1' function as a reference. The 'wakeup1' function wakes up all processes sleeping on a given 'chan'. It loops through the process table ('ptable.proc'), checks if each process is in the 'SLEEPING' state and if its channel matches the given 'chan', and sets its state to 'RUNNABLE'.

For the 'wakeup_single' function, you can simply call the 'wakeup1' function with the given 'chan' argument since it already wakes up all the relevant processes. The 'wakeup_single' function can be defined as:

void wakeup_single(void* chan) {
wakeup1(chan);
}