基本上，我正在为实现多级优先级队列的 xv6 内核实现一个调度程序。我有一个严重的问题，我不明白，我类(class)的助教不明白，我已经错过了这个项目的最后期限，所以现在帮助我不会给我任何加分 - 但我想知道为什么我有以下行为......
首先，这是我为 xv6 更改的原始调度程序(为了比较 - 这不是我的实现):
// Per-CPU process scheduler.
// Each CPU calls scheduler() after setting itself up.
// Scheduler never returns. It loops, doing:
// - choose a process to run
// - swtch to start running that process
// - eventually that process transfers control
// via swtch back to the scheduler.
void
scheduler(void)
{
struct proc *p;
for(;;){
// Enable interrupts on this processor.
sti();
// Loop over process table looking for process to run.
acquire(&ptable.lock);
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
if(p->state != RUNNABLE)
continue;
// Switch to chosen process. It is the process's job
// to release ptable.lock and then reacquire it
// before jumping back to us.
proc = p;
switchuvm(p);
p->state = RUNNING;
swtch(&cpu->scheduler, proc->context);
switchkvm();
// Process is done running for now.
// It should have changed its p->state before coming back.
proc = 0;
}
release(&ptable.lock);
}
}
新的调度器的思路是这样的:在ptable里面有一个数组，里面填充了proc结构体。我将此过程数组中的每个元素称为“p”，它们包含基本信息(例如它们拥有的“门票”数量或状态等)。我需要在一个时间片内运行所有高优先级 (HP) p，然后将它们的优先级更改为低。当没有 HP 过程时，我“随机”选择一个 LP 过程并运行它两个时间片。我的算法如下:
scheduler()
for(;;) //scheduler NEVER completes
//information gathering
for (entire proc array) //goes over it once
gather how many HP and LP procs
count total HP and LP tickets in each proc (for lottery)
if #HP > 1 //randomly choose a HP proc
hold HP lottery, run one HP proc //afterwards,
int rand = random() % num_HP_tickets
for (entire array)
curr_index_of_tickets += p->num_tickets;
if (curr_index_of_tickets >= rand) //we found the right p!
run p for one time slice
else if #HP == 1
find and run the one HP proc
else if #HP < 1 //then no HP procs! Time to run LP
if #LP > 1
hold LP lottery, run one LP proc for two time slices, similar to above
else if #LP == 1
find the LP proc, run it
这就是问题所在...我的 proc 似乎总是等于 0。它从来没有看到 p 中的信息，也没有从 p 收集 proc 信息，等等。我不知道为什么。
我用大量的打印输出语句进行了测试。我会先在这里发布它的输出:
entered scheduler
entered INFORMATION loop: iteration 0
Proc is 0
Proc is 0 even after proc gets p from for loop
Information 0: found 0 HP procs
Information 0: found 0 LP procs
Num HP: 0
Num LP: 0
Num HP Tickets: 0
Num LP Tickets: 0
This concludes loop #: 0
entered INFORMATION loop: 1
Proc is 0
entered INFORMATION loop: 2
Proc is 0
entered INFORMATION loop: 3
Proc is 0
etc.....
再一次，不知道为什么这不起作用...我确定有多个错误，而且我有一大堆打印出来的语句只是为了看看哪里出了问题。这也需要相当多的调试工作，所以我对是否有人有答案我不太乐观......为此目的，以及这些警告，这是我的整个调度程序功能。抱歉，长度...
void
scheduler(void)
{
struct proc *p;
for(;;){
//TODO: Remove statement
cprintf("entered scheduler\n");
// Enable interrupts on this processor.
sti();
// Loop over process table looking for process to run.
acquire(&ptable.lock);
//keeps track of number of procs (to be used for lottery RNG)
int num_LP_t = 0;
int num_HP_t = 0;
int num_HP = 0;
int num_LP = 0;
int rand = 0;
int curr_tickets = 0;
//goes through once to complete information gathering for HP and LP queue
//TODO: remove i - for testing only
int i = -1;
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
i++;
//TODO: Remove statement
cprintf("entered INFORMATION loop: %d\n", i);
if (proc == 0) cprintf("Proc is 0\n");
if(p->state != RUNNABLE)
continue;
// Switch to chosen process. It is the process's job
// to release ptable.lock and then reacquire it
// before jumping back to us.
//TODO: uncomment? proc = p;
if (proc == 0) cprintf("Proc is 0 even after proc gets p in INFORMATION LOOP\n");
cprintf("Information %d: found %d HP procs\n", i, num_HP);
cprintf("Information %d: found %d LP procs\n", i, num_LP);
cprintf("Num HP: %d\n", num_HP);
cprintf("Num LP: %d\n", num_LP);
cprintf("Num HP Tickets: %d\n", num_HP_t);
cprintf("Num LP Tickets: %d\n", num_LP_t);
cprintf("This concludes loop #: %d\n", i);
if (p->priority_level == 1){
num_HP++;
num_HP_t += p->num_tickets;
}
if (p->priority_level == 0){
num_LP++;
num_LP_t += p->num_tickets;
}
}//end information loop
cprintf("Begin HP Queue:\n");
if (num_HP > 1){
cprintf("HP Queue had: %d procs to run\n", num_HP);
rand = random() % num_HP_t;
cprintf("We choose our random to be: %d\n", rand);
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
if (p->state != RUNNABLE) continue;
//TODO: uncomment? proc = p;
if (proc == 0) cprintf("Proc is 0 even after setting proc = p in HP queue\n");
if (p->priority_level == 1){
cprintf("Found a HP Proc while searching for rand, currently at: %d\n", curr_tickets);
curr_tickets += p->num_tickets;
if (curr_tickets >= rand){
if (proc == 0)
cprintf("proc is 0 while in HP queue 1\n");
//proc = p;
switchuvm(p);
p->state = RUNNING;
/*cprintf("Proc Info:");
cprintf("uint sz: %d", proc->sz);
cprintf("enum procstate state: %s", proc->state);
cprintf("volatile int pid: %d", proc->pid);
cprintf("int killed: %d", proc->killed);
cprintf("int priority_level: %d", proc->priority_level);*/
swtch(&cpu->scheduler, proc->context);
switchkvm();
if (p->state == RUNNABLE){
cprintf("HP process still runnable after 1 TS.\nHere is the updated information\n");
p->priority_level = 0;
num_HP--;
num_LP++;
num_HP_t -= p->num_tickets;
num_LP_t += p->num_tickets;
cprintf("Num HP: %d\n", num_HP);
cprintf("Num LP: %d\n", num_LP);
cprintf("Num HP Tickets: %d\n", num_HP_t);
cprintf("Num LP Tickets: %d\n", num_LP_t);
}
}
}
}
}
else if (num_HP == 1){
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
if (p->state != RUNNABLE) continue;
//TODO: Uncomment? proc = p;
if (proc == 0) cprintf("Proc is 0 even after setting proc = p in HP == 1 queue\n");
if (p->priority_level == 1){
//proc = p;
switchuvm(p);
p->state = RUNNING;
swtch(&cpu->scheduler, proc->context);
switchkvm();
if (p->state == RUNNABLE){
p->priority_level = 0;
num_HP--;
num_LP++;
num_HP_t -= p->num_tickets;
num_LP_t += p->num_tickets;
}
}
}
}//end else num_HP = 1
else if (num_HP < 1){
if (num_LP > 1){
rand = random() % num_LP_t;
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
if (p->state != RUNNABLE) continue;
//TODO: uncomment? proc = p;
if (proc == 0) cprintf("Proc is 0 even after setting proc = p in LP > 1 queue\n");
if (p->priority_level == 0){
curr_tickets += p->num_tickets;
if (curr_tickets >= rand) {
//proc = p;
switchuvm(p);
p->state = RUNNING;
swtch(&cpu->scheduler, proc->context);
switchkvm();
if (p->state != RUNNABLE){
num_LP--;
num_LP_t -= p->num_tickets;
}
}
}
}
}
else if (num_LP == 1){
for(p = ptable.proc; p < &ptable.proc[NPROC]; p++){
if (p->state != RUNNABLE) continue;
//TODO: Uncomment? proc = p;
if (proc == 0) cprintf("Proc is 0 even after setting proc = p in LP == 1 queue\n");
if (p->priority_level == 0){
//proc = p;
switchuvm(p);
p->state = RUNNING;
swtch(&cpu->scheduler, proc->context);
switchkvm();
if (p->state != RUNNABLE){
num_LP_t -= p->num_tickets;
}
}
}
}
}//end lp
proc = 0;
release(&ptable.lock);
}//end outer for loop (;;)
}// end scheduler
最佳答案
注释掉 proc=p。显而易见的事情是显而易见的。
关于c - xv6 调度程序 - proc 从不!= 0，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22006642/