9608_s20_qp_12
A paper of Computer Science, 9608
Questions:
7
Year:
2020
Paper:
1
Variant:
2

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The steps 1 to 6 describe the first pass of a two‑pass assembler. The following three statements are used to complete the sequence of steps. A If it is already in the symbol table, it checks to see if the absolute address is known B When it meets a symbolic address, it checks to see if it is already in the symbol table C If it is known, it is entered Write one of the letters A, B or C in the appropriate step to complete the sequence. 1. The assembler reads the assembly language instructions 2. 3. If it is not, it adds it to the symbol table 4. 5. 6. If it is not known, it is marked as unknown. The assembler translates assembly code into machine code. The table shows the denary values for three assembler op codes. Op code Denary value LDD ADD STO Convert the denary value for the op code LDD into 8‑bit binary. Convert the denary value for the op code STO into hexadecimal. State why the denary value for the op code ADD cannot be represented in 8‑bit two’s complement form. Justify your answer. The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (. Instruction Explanation Op code Operand LDM #n Immediate addressing. Load the denary number n to ACC LDD Direct addressing. Load the contents of the location at the given address to ACC LDX Indexed addressing. Form the address from + the contents of the Index Register. Copy the contents of this calculated address to ACC LDR #n Immediate addressing. Load the denary number n to IX STO Store contents of ACC at the given address ADD Add the contents of the given address to ACC INC Add 1 to the contents of the register (ACC or CMP Compare contents of the address given with the contents of ACC JPE Following a compare instruction, jump to if the compare was True JPN Following a compare instruction, jump to if the compare was False JMP Jump to the given address OUT Output to screen the character whose ASCII value is stored in ACC END Return control to the operating system
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A software development company has a relational database, SOFTWARE_MANAGEMENT. The database stores details of the customers who have purchased software, as well as the software and licences that customers have purchased. The SOFTWARE_MANAGEMENT database has the following tables: CUSTOMER_DETAILS(CustomerID, CompanyName, Address1, Address2, City) SOFTWARE_PURCHASED( SoftwareName, SoftwareDescription, CustomerID, LicenceType, LicenceCost, RenewalDate) Explain why this database is not in Third Normal Form (3NF). Refer to the tables in your answer. Do not attempt to normalise the tables. Give an example from the database SOFTWARE_MANAGEMENT for each of the following database terms. Term Example Entity Foreign key Attribute The company also develops computer games. They extend the relational database SOFTWARE_MANAGEMENT by adding a new table. The new table, GAME_DEVELOPMENT, stores details about the games and the software development teams creating them. The table shows example data in GAME_DEVELOPMENT. GameName Genre TeamNumber DevelopmentStage ManagerID Bunny Hop Platform Analysis 23KP Fried Eggs Retro Programming stage 1 9RTU Create‑a‑game Action Acceptance testing 11TF Complete the Data Definition Language (DDL) statement to create the table GAME_DEVELOPMENT. CREATE ………………………………………………………… ………………………………………………………… ( GameName VarChar, Genre VarChar, ………………………………………………………… ………………………………………………………… , DevelopmentStage VarChar, ManagerID VarChar, ………………………………………………………… (GameName) ); Another table, PRODUCT_MANAGER, is created. PRODUCT_MANAGER(ManagerID, FirstName, LastName) Complete the Data Manipulation Language (DML) statement to return the game name, genre and team number of all games managed by the product manager with the first name ‘James’ and the last name ‘Fitz’. ………………………………………………………… GameName, Genre, TeamNumber FROM GAME_DEVELOPMENT, PRODUCT_MANAGER WHERE PRODUCT_MANAGER.FirstName = "James" AND PRODUCT_MANAGER.LastName = "Fitz" AND ………………………………………………………………………………………………… = …………………………………………………………………………………………………… ;
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