10110; BR = 11100111; CR = 10110001; DR = 10111010 Determine the 8-bit values in each register after the execution of the following sequence of microopera
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The 8-bit register AR, BR, CR, and DR initially have the following values: [5]
AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010
Determine the 8-bit values in each register after the execution of the following sequence
of microoperations.
AR AR + BR Add BR + AR
CR CR AND DR, BR BR + 1 AND DR to CR, Increment BR
AR AR - CR Subtract CR from AR
Step by step
Solved in 2 steps
- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?The 8-bit register AR, BR, CR, and DR initially have the following values: [5]AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010Determine the 8-bit values in each register after the execution of the following sequence of microoperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from ARThe 8-bit register AR, BR, CR, and DR initially have the following values:AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010Determine the 8-bit values in each register after the execution of the following sequence of microoperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from AR
- Q2. The 8-bit registers AR, BR, CR, and DR initially have the following values: AR= 11010010, BR= 11100011, CR= 10111001 , DR= 10101011 Determine the 8-bit values in each register after the execution of the following sequence of microoperations. AR <-AR + BR CR <- CR ^ DR, BR <- CR + 1 AR <-AR – CRThe 8-bit register AR, BR, CR, and DR initially have the following values:AR = 11010010; BR = 11111111; CR = 10101001; DR = 10101010Determine the 8-bit values in each register after the execution of the following sequence ofmicrooperations.AR <- AR + BR Add BR + ARCR <- CR AND DR, BR <- BR + 1 AND DR to CR, Increment BRAR <- AR - CR Subtract CR from ARThe 8-bit register AR, BR, CR, and DR initially have the following values: AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010 Determine the 8-bit values in each register after the execution of the following sequence of microoperations. AR AR + BR Add BR + AR CR CR AND DR, BR BR + 1 AND DR to CR, Increment BR AR AR - CR Subtract CR from AR
- The 8-bit registers R1, R2, R3, and R4 initially have the following values: R1 - 1111 0010, R2- 1 1 1 1 1 1 1 1 , R3- 1011 1001 , R4 1 1 101010 Determine the 8-bit values in each register after the execution of the following sequence of microoperations. R1 <-R1 + R2 R3<- R3 ^ R4, R2 <- R2 + 1 R1 <- R1 R33 The 8-bit registers R1, R2, R3, and R4 initially have the following values:R1 - 1111 0010, R2- 1 1 1 1 1 1 1 1 , R3- 1011 1001 , R4 1 1 101010Determine the 8-bit values in each register after the execution of the following sequence ofmicrooperations.R1 <-R1 + R2R3<- R3 ^ R4, R2 <- R2 + 1R1 <- R1 – R3Consider the following two operations: (a) PC = PC + 4; (b) PC = PC + offset (offset is a positive or negative multiple of 4).Which of the following statements is true? a In the MIPS architecture, operation (b) is done by performing operation (a) as many times as required. b Operation (a) requires about 1/2 as much hardware to implement as does operation (b) c Operation (a) is simply done by replacing the right-most two bits of PC with 00. d Since (a) is a special case of (b), these two operations require the same hardware to implement.
- 1. Suppose 8 bit registers have following contentsX=00001111Y=10101010Z= 11011011W=00110011What will be the 8 bit values of each register after execution of following sequences ofmicrooperations ?X ← ? + ?Z←Z⋀ ?, ? ← ? + 1X←X-ZORIG x3200 LDI R1, num1 LD R2, num2 JSR NAND LEA R5, num1 STR R3, R5, #3 HALT NAND AND R3, R2, R1 NOT R3, R3 RET BASE .FILL x3300 num1 .FILL x3301 num2 .FILL x3302 .END Contents of memory: x3300 1010 1010 0011 1000 x3301 1111 0000 1100 0000 x3302 0101 1010 0001 0000 x3303 1111 1111 0000 0000 Find Content of R2 at the line of code LD R2, num2; in the above assembly language code:.ORIG x3200 LDI R1, num1 LD R2, num2 JSR NAND LEA R5, num1 STR R3, R5, #3 HALT NAND AND R3, R2, R1 NOT R3, R3 RET BASE .FILL x3300 num1 .FILL x3301 num2 .FILL x3302 .END Contents of memory: x3300 1010 1010 0011 1000 x3301 1111 0000 1100 0000 x3302 0101 1010 0001 0000 x3303 1111 1111 0000 0000 Find Content of R1 at the line of code LDI, R1, num1; in the above assembly language code: Group of answer choices xF0C0 x320A x3301 x3200