Suppose we are in following code, at line 4. --What is the entry at the top of k

Question

Suppose we are in following code, at line 4.
--What is the entry at the top of kernel stack A (old) ?
--What is the next entry?
--And the next ?
--And the next few entries (their order is not clearly known from the figures) but you can say what they are as a group.
1 # void swtch(struct context **old, struct context *new); 2 # 3 # Save current register context in old 4 # and then load register context from new. 5 .globl swtch 6 swtch: 7 # Save old registers 8 movl 4(%esp), %eax # put old ptr into eax 9 popl 0(%eax) # save the old IP 10 movl %esp, 4(%eax) # and stack 11 movl %ebx, 8(%eax) # and other registers 12 movl %ecx, 12(%eax) 13 movl %edx, 16(%eax) 14 movl %esi, 20(%eax) 15 movl %edi, 24(%eax) 16 movl %ebp, 28(%eax) 17 18 # Load new registers 19 movl 4(%esp), %eax # put new ptr into eax 20 movl 28(%eax), %ebp # restore other registers 21 movl 24(%eax), %edi 22 movl 20(%eax), %esi 23 movl 16(%eax), %edx 24 movl 12(%eax), %ecx 25 movl 8(%eax), %ebx 26 movl 4(%eax), %esp # stack is switched here 27 pushl 0(%eax) # return addr put in place 28 ret # finally return into new ctxt

Suppose we are in following code, at line 4.
--What is the entry at the top of kernel stack A (old) ?
--What is the next entry?
--And the next ?
--And the next few entries (their order is not clearly known from the figures) but you can say what they are as a group.
1 # void swtch(struct context **old, struct context *new); 2 # 3 # Save current register context in old 4 # and then load register context from new. 5 .globl swtch 6 swtch: 7 # Save old registers 8 movl 4(%esp), %eax # put old ptr into eax 9 popl 0(%eax) # save the old IP 10 movl %esp, 4(%eax) # and stack 11 movl %ebx, 8(%eax) # and other registers 12 movl %ecx, 12(%eax) 13 movl %edx, 16(%eax) 14 movl %esi, 20(%eax) 15 movl %edi, 24(%eax) 16 movl %ebp, 28(%eax) 17 18 # Load new registers 19 movl 4(%esp), %eax # put new ptr into eax 20 movl 28(%eax), %ebp # restore other registers 21 movl 24(%eax), %edi 22 movl 20(%eax), %esi 23 movl 16(%eax), %edx 24 movl 12(%eax), %ecx 25 movl 8(%eax), %ebx 26 movl 4(%eax), %esp # stack is switched here 27 pushl 0(%eax) # return addr put in place 28 ret # finally return into new ctxt

Suppose we are in following code, at line 4.
--What is the entry at the top of kernel stack A (old) ?
--What is the next entry?
--And the next ?
--And the next few entries (their order is not clearly known from the figures) but you can say what they are as a group.
1 # void swtch(struct context **old, struct context *new); 2 # 3 # Save current register context in old 4 # and then load register context from new. 5 .globl swtch 6 swtch: 7 # Save old registers 8 movl 4(%esp), %eax # put old ptr into eax 9 popl 0(%eax) # save the old IP 10 movl %esp, 4(%eax) # and stack 11 movl %ebx, 8(%eax) # and other registers 12 movl %ecx, 12(%eax) 13 movl %edx, 16(%eax) 14 movl %esi, 20(%eax) 15 movl %edi, 24(%eax) 16 movl %ebp, 28(%eax) 17 18 # Load new registers 19 movl 4(%esp), %eax # put new ptr into eax 20 movl 28(%eax), %ebp # restore other registers 21 movl 24(%eax), %edi 22 movl 20(%eax), %esi 23 movl 16(%eax), %edx 24 movl 12(%eax), %ecx 25 movl 8(%eax), %ebx 26 movl 4(%eax), %esp # stack is switched here 27 pushl 0(%eax) # return addr put in place 28 ret # finally return into new ctxt 1 # void swtch(struct context **old, struct context *new); 2 # 3 # Save current register context in old 4 # and then load register context from new. 5 .globl swtch 6 swtch: 7 # Save old registers 8 movl 4(%esp), %eax # put old ptr into eax 9 popl 0(%eax) # save the old IP 10 movl %esp, 4(%eax) # and stack 11 movl %ebx, 8(%eax) # and other registers 12 movl %ecx, 12(%eax) 13 movl %edx, 16(%eax) 14 movl %esi, 20(%eax) 15 movl %edi, 24(%eax) 16 movl %ebp, 28(%eax) 17 18 # Load new registers 19 movl 4(%esp), %eax # put new ptr into eax 20 movl 28(%eax), %ebp # restore other registers 21 movl 24(%eax), %edi 22 movl 20(%eax), %esi 23 movl 16(%eax), %edx 24 movl 12(%eax), %ecx 25 movl 8(%eax), %ebx 26 movl 4(%eax), %esp # stack is switched here 27 pushl 0(%eax) # return addr put in place 28 ret # finally return into new ctxt

Explanation / Answer

No, its not possible

Let two process be P1, P2

Let three resources be R1, R2, R3

Case 1: P1 and P2 needs <=1 resources each.

no deadlock because we have 3 resources whereas we need max. 2 resources total

Case 2: P1 needs 1 and P2 needs 2 resources (or vice versa)

no deadlock since we have 3 resources and we need 3 resources.

Case 3: P1 and P2 both need 2 resources.

no deadlock again because we have 3 resources and we need max. 4 hence at a time only one process will be waiting for the resource to get free and when only one process is waiting then deadlock does not arise. (for a deadlock to happen at least 2 process should be waiting at the same time).

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