Practice Problem 3.1 (solution page 361) Assume the following values are stored at the indicated memo registers: Address Value 0x100 OxFF 0x104 OxAB 0x108 0x13 0x10C 0x11 Operand %rax 0x104 $0x108 (%rax) 4 (%rax) 9(%rax, %rdx) Register %rax %rcx %rdx Fill in the following table showing the values for the indicat Operand %rax 0x104 $0x108 (%rax) 4 (%rax) 9 (%rax, %rdx) 260(%rcx, %rdx) OxFC (,%rcx, 4) (%rax,%rdx, 4) 260 (%rcx, %rdx) OXFC (,%rcx, 4) (%rax, %rdx, 4) Value 0x100 0x1 0x3 Value Value Comment 0x100 Register OxAB 0x108 OxFF OxAB 0x11 0x13 OxFF 0x11 Absolute address Immediate Address 0x100 Address 0x104 Address 0x10C Address 0x108 Address 0x100 Address 0x10C

Practice Problem 3.1 (solution page 361) Assume the following values are stored at the indicated memo registers: Address Value 0x100 OxFF 0x104 OxAB 0x108 0x13 0x10C 0x11 Operand %rax 0x104 $0x108 (%rax) 4 (%rax) 9(%rax, %rdx) Register %rax %rcx %rdx Fill in the following table showing the values for the indicat Operand %rax 0x104 $0x108 (%rax) 4 (%rax) 9 (%rax, %rdx) 260(%rcx, %rdx) OxFC (,%rcx, 4) (%rax,%rdx, 4) 260 (%rcx, %rdx) OXFC (,%rcx, 4) (%rax, %rdx, 4) Value 0x100 0x1 0x3 Value Value Comment 0x100 Register OxAB 0x108 OxFF OxAB 0x11 0x13 OxFF 0x11 Absolute address Immediate Address 0x100 Address 0x104 Address 0x10C Address 0x108 Address 0x100 Address 0x10C

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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Problem Question 03 (CO3) [0.5 + 0.5 = 1]: Given below are the contents of several Intel 8086 registers and PHYSICAL memory addresses (ALL in hexadecimal): Registers: Memory Locations [Physical Address] = Contents [05000] = 3300 [06000] = 4444 [07000] = 5555 [95000] = 367A [96000] = 6666 [97000] = 10C5 DI = 3000 %3D BX = 3000 %3D ВР 3 С345 For the following instructions, determine the contents of AX after the each of the instruction has been executed: (а) MOV (b) MOV АХ, ВР АХ, [ВХ+DI] AX = AX
[1] ( Show your work. Show hoe you compute memory address by using the effective memory address computation. Assume the following values are stored at the indicated memory addresses and registers: Address Value 0x100 OxFF 0x104 OxAB 0x108 0x13 0x10c 0x11 Register %rax %rcx %rdx $0x108 (%rax) 4(%rax) 9(%rax, %rdx) 260(%rcx,%rdx) OxFC (,%rcx, 4) (%rax, %rdx, 4) Value 0x100 0x1 0x3 Fill in the following table showing the values for the indicated operands: Operand Value %rax 0x104
3. Show the complied MIPS code for the following C code, assume x and y is in $s1 and $s2 respectively, and the base address of A is in $s3. 2 y = x + A[5] + A[4] |+1 A[6] = A[6] - y
3. Consider the following AT&T ISA 32 ASM codes for a C for loop: loop: puahl tabp tenp, tebp 8 (sebp), ecx 12 (tebp), tedx $0, teax Becx, tedx movl movl movl movl cmpl jge .L3 .L6: subl addl addl cmpl jg $1, tecx $1, Bedx $1, teax Sedx, tecx .L6 L3: $1, teax Sebp addl popl ret Based on the assembly code above, (Note: you may only use the symbolic variables x, y, and result in your expressions below – do not use register names.) int loop (int x, int y) int result; result++ ) { for return result;
3- What is the difference between: a- MOV BX,[1234H] and LEA BX,[1234H] b- LDS AX,[200H] and LES AXX,[200H] 4-Use MOV to load address of memory MEM1.
Problem 1: Consider the following assembly code: .LCO .string "ans %d\n" main: .LFBO: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LCO, %edi movl $0, %ebx call printf leave ret This code came from the following skeleton C file, and optimized with 00, so gcc -00 -S was the exact command to compile the file. Complete the below given C code using the above information. Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file. #include int main() { long int val= int result= for(;val>. val=. result += printf("ans %d\n",result); }
[4.3] Consider the following code: 1h x2, 0 (x10) sw x2, 8 (x10) Assume that the register x10 contains the address 0x80001000 and the data at address is OxABCDEF00. a. Write the required instructions to add the address 0x80001000 into the register x10? b. What value is stored in 0x80001008 on a little-endian machine?
Q1: Determine the physical address of the source operand base on the Based Indexed Addressing Mode. The MOV instruction MOV AX, [BX].[16A0] [SI] The contents of IP, CS and DS are 0120, A342p, and 2C60, respectively. Also, the contents of BX and SI are 6752, and C344, respectively. Explain by the draw the registers and the logical of the system memory in the before execution and after execution. Note: the content of PA is 2B7C and the coding of this instruction is XXXX
D] An integer arithmetic unit that can perform addition and multiplication of 16-bit unsigned numbers is to be used to multiply two 32-bit unsigned numbers. All operands, intermediate results, and final results are held in 16-bit registers labeled R0 through R15. The hardware multiplier multiplies the contents of Ri (multiplicand) by Rj (multiplier) and stores the double-length 32-bit product in registers Rj and Rj+1, with the low-order half in Rj. When j = i − 1, the product overwrites both operands. The hardware adder adds the contents of Ri and Rj and puts the result in Rj. The input carry to an Add operation is 0, and the input carry to an Add-with-carry operation is the contents of a carry flag C. The output carry from the adder is always stored in C. Specify the steps of a procedure for multiplying two 32-bit operands in registers R1, R0, and R3, R2, high-order halves first, leaving the 64-bit product in registers R15, R14, R13, and R12. Any of the registers R11 through R4 may…
Problem 2.2: write down the sequence of the instruction (till HLT instruction) if the program begin with the location 1FFH using two scenario : 1) data =00, 2) data-01 Address (Hex) IFF5 8085 Instruction XRA LXI IFFB IFF6 IFF9 PCHL IFFA HLT LXI H,2100 H ANI Data LXI H, FFFF H IFFB IFFE 2000 2003 2004 INX H JZ LOOP HLT 2103 LXI H, IFFF H MOV A,M INR A LOOP 2104 2105 2106 HLT
[5.2] Consider the following code: 1b x6, 0 (x7) sw x6, 4(x7) Assume that the register x7 contains the address Ox10000000 and the data at address is 0x11223344. ايتل a. What value is stored in 0x10000004 on a big-endian machine? b. What value is stored in 0x10000004 on a little-endian machine?
2. Consider the following MIPS code segment being executed on machine M: lw $t1, 10(St2) srl $t1, $t1, 16 lui $13 0x7FFF ori $t3 0xFFFF Loop: xor $14, $t1, St3 add $t1, $t1, St4 sll $t1, $t1, 16 bgt $t1, Szero, Loop sw $11, 1000(S13) Assume M has a clock frequency is 3.5 GHz. Also, assume that 1w needs 5 clock cycles, sv 4 clock cycles, bne 2 clock cycles, and all other instructions need 3 clock cycles to execute. Calculate the total amount of CPU-time (in nanoseconds) required to execute the above code.