Postgraduate entrance exam 408 includes data structure, computer network, computer composition principle and operating system four subjects, each of which is the content that programmers should focus on understanding and mastering.

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Analysis is my own writing, if there are questions or mistakes please comment to inform, hope to point a big praise 💗

The data structure

The node construction of bidirectional circular linked list is known as shown in the figure. The process of inserting the node with pointer P after the node pointed to by pointer Q in the linked list is executed successively ___ (Beihang University 2010)

A. p->llink=q; p->rlink=q->rlink; q->rlink=p; q->llink=p;
B. p->llink=q; p->rlink=q->rlink; q->rlink=p; q->rlink->llink=p;
C. p->llink=q; p->rlink=q->rlink; q->rlink->llink=p; q->rlink=p;
D. p->llink=q; p->rlink=q->rlink; q->rlink=p; p->llink->llink=p;
Copy the code

Answer and analysis

Answer: C

🔊 Insert elements into a bidirectional circular list (order must be present, otherwise the link will be broken) :

  1. First, the left and right Pointers of Q point to the successors of P and P respectively
  2. Then point back to the left pointer of the subsequent element of P
  3. And finally, I’m going to point to the right pointer of P
p->llink=q; p->rlink=q->rlink; (1)
q->rlink->llink=p; (2)
q->rlink=p; (3)
Copy the code

Computer network

The routing algorithm used by RIP is ___ (Chongqing University of Posts and Telecommunications 2007)

A. Distance vector algorithm B. link-state algorithm C. Neighbor test message pair algorithm D. None of the aboveCopy the code

Answer and analysis

Answer: A,

🔊 RIP is based on the distance vector algorithm, and OSPF is based on the link-state algorithm.

Computer composition principle

A machine word length of 64 bits, memory capacity of 256MB, such as word addressing, its addressing range is ____ (Beihang University 2015)

A. 0 ~ 8M-1
B. 0 ~ 16M-1
C. 0 ~ 32M-1
D. 0 ~ 64M-1
Copy the code

Answer and analysis

Answer: C

🔊 the length of the machine word is 64 bits, that is, 8 bytes. The address by word needs to be 256MB/8B = 32M. Therefore, C is selected.

The operating system

If a system has 64MB of memory and a 32-bit address processor, its virtual address space is ___ bytes (Nanjing University of Science and Technology 2011)

A. 2GB
B. 4GB
C. 100KB
D. 64GB
Copy the code

Answer and analysis

Answer: B

🔊 the virtual address space is based on the number of processor addresses. In this case, it can be 2 to the power of 32, that is, 4GB.