EBK DATABASE CONCEPTS
EBK DATABASE CONCEPTS
7th Edition
ISBN: 9780133544886
Author: AUER
Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)
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Chapter 6, Problem 6.34RQ
Program Plan Intro

Cursor:

The term cursor is used to retrieve the information in row by row manner. If the records stored in the database table, which needs to be updated one row at a time is carried over by cursor.

  • It plays the major role for fetching a required row from the table in the database which holds numerous records.
  • Cursors are defined using SELECT statement.
  • It points the set of rows which is resulted from an SQL SELECT statement.
  • In general the cursor is placed in the first or last row.
  • It influences the performance of the SQL Server as it uses the SQL server instances memory, decrease network bandwidth, and reduction in concurrency.
  • It’s recommended to avoid the use of cursor and the cursor can be replaced with WHILE loop, temporary tables, and sub queries.
  • The types of cursor are listed below:
    • Forward-only cursor
    • Static cursor
    • Dynamic cursor
    • Keyset cursor

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Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction lw St1, 48($10) What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Data Memory Select one a. Ox00001724 Ob. 0x00002DE4 Oc 0x000016F4 Od 0x00001720 Oe. 0x00002DEO Clear my choice Rea " Address Content 0x000016EC 0x0000ABOD 0x000016F0 0x00A01245 0x000016F4 0x00001A42 0x000171C 0x0008124F 0x0001720 0x00021345 0x0001724 0x000067AB ALU Ox0001734 0x0000AB35 0x0001738 0x0000FA72 0x0000ABOC ALS 0x000174C Register File and PC $to = 0x000016FO $t1 = 0x000016F4 PC = 0x0000148A0 (Before executing LW)
Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction SLL S10, $3,4 What is the value of the element pointed by arrow number 1 by in hexadecimal or binary? Note that the PC and the content of registers are found in bottom left of the figure below Address Select one Oa. Ox00000000 Ob. Ob 10011' Oc 0x0000AF00 ® d. 06/01000 Oe. None of the options Of Ox000000BA LEY Content 0x000016EC Dx0000ABOD Data Memory 0x000016F0 0x00A01245 0x000016F4 0x00001A42 Ox000171C 0x0008124 0x0001720 0x00021345 0x0001724 0x000067AB 0x0001734 0x0001738 0x0000A835 0x0000FA72 0x000174C 0x0000AC $63 0x000000BA PC = 0x000FAC04 Register File and PC (Before executing SLL)
Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction lw $10, 48(S1) What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Select one a. None of the options Ob 0x00014C64 Oc 0x00014CC4 Od 0x00014BA4 Oe. Ox14C1C Of 0x00014C60 Og 0x00014CCO D Data Memory Address Content 0x000016EC 0x0000ABOD 0x000016F0 0x00A01245 0x000016F4 0x00001A42 0x000171C 0x0008124F 0x0001720 0x00021345 0x0001724 0x000067AB 0x0001734 0x0000AB35 0x0001738 0x0000FA72 0x000174C 0x0000ABOC ALU Register File and PC Sto Ox000016F0 St1 = 0x000016F4 PC = 0x00014BA0 (Before executing LW).
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