of empty blocks in a linked list in the order of their address, where each node stores the size of next empty plock in the memory. Say that current state of the memory is hold by the linked list shown below, which tells that there are currently four empty blocks in the memory, and the size of first block is 12 byte etc. 12 8 3 Assume that following memory block requests have been made in given order: 8 bytes, 2 bytes and 6 bytes. Draw the linked list to be changed after each block assignment by the following strategies; (a) First Fit (a) Best Fit

of empty blocks in a linked list in the order of their address, where each node stores the size of next empty plock in the memory. Say that current state of the memory is hold by the linked list shown below, which tells that there are currently four empty blocks in the memory, and the size of first block is 12 byte etc. 12 8 3 Assume that following memory block requests have been made in given order: 8 bytes, 2 bytes and 6 bytes. Draw the linked list to be changed after each block assignment by the following strategies; (a) First Fit (a) Best Fit

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider a Single Linked List (SLL) of nodes, where ‘node’ is a type with two fields: integer data and the node type reference ‘next’ that refers to the next node in the list. For each part given below consider the SLL and the code given in the second column. Following the execution of the code, show the resulting list (in the answer box, for each part, use the arrow key to draw the list as: head-----> ? -------> ? --------> ?). List and Java Code a head ⟶ 1⟶ 2⟶ 3⟶ 4⟶ null node ptr=head; node newnode = new node(0); // data=0 head = newnode; newnode.next=ptr; b head ⟶ 1⟶ 2⟶ 3⟶ 4⟶ null node ptr=head; node newnode = new node(6); // data=6 while(ptr.next!=null) { ptr=ptr.next; } ptr.next=newode; c head ⟶ 1⟶ 2⟶ 3⟶ 4⟶ null node ptr=head; ptr=ptr.next; head=ptr;
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Consider a Single Linked List (SLL) of nodes, where 'node' is a type with two fields: integer data and the node type reference 'next' that refers to the next node in the list. For each part given below consider the SLL and the code given in the second column. Following the execution of the code, show the resulting list (in the answer box, for each part, use the arrow key to draw the list as: head-----> ? -------> ? -------> ?). List and Java Code | head →1-2-3→4→ null node ptr=head; node newnode = new node(0); // data=0 head = newnode; newnode.next=ptr; b head 1→2→3→4→ null node ptr=head; node newnode = new node(6); // data=6 while(ptr.next!=null) { ptr=ptr.next; } ptr.next=newode; head -1-2- 3→ 4→ mull node ptr=head; ptr=ptr.next; head=ptr;
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Q-3-2 Given the following Linked list HEAD 25 Apply the following operations on the above Linked List by drawing the content of the linked list after each operation: addEnd(5), addEnd(10), addEnd (33), deleteHead(25), deleteHead(5), addEnd (55), addEnd(14), addEnd(18), deleteHead(33), addEnd(100)
Objective Define a circular buffer data structure and test it. Problem Description: A circular buffer (also called a circular queue) is circular list of nodes where data items are added on one end of the buffer and removed from the other end. Because the nodes form a circular list, the list has no end or beginning: the tail node points to the head node, creating a ring of nodes. You may think of the nodes as containers or slots that are all initially empty but can be assigned a value in their data field info. Every time a new data item is inserted (inserting to a buffer is often referred as Writing), one slot is filled and the buffer has one less empty slot. Every time a data item is removed (referred to as Reading), the buffer has one more empty slot. Since the list has no beginning and no end, a pointer (writeIndex) is used to mark the next empty slot to write to and a second pointer (readIndex) is used to mark the next node to read from. The readIndex/writeIndex must be…
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Memory Allocation (20 minutes) Assume you have a buddy allocator that initially has a single 512 byte free block. You cannot allocate more memory. The first allocation is for 200 bytes. Describe how the allocation occurs with the buddy algorithm. State what's in the free list after the allocation. How many bytes are lost to internal fragmentation after this first allocation? Next, we deallocate the 200 byte allocation. Describe what happens. State what the free list looks like now. Now, we get 4 different 100 byte allocations. You don't have to describe the process. However, state the free list at the end of all 4 allocations. How many bytes are lost due to internal fragmentation after the 4 allocations? Describe an approach you could take to fit another 100 byte allocation within your original 512 bytes. You can assume you can use your approach to allocate all 5 100 byte allocations from scratch.