Problem 7.4 When given the C code 1 void cond(int a, int *p) 2 { 3 4 if (p && a > 0) *p += a; a MIPS compiler generates the following assembly code for the body of the instruction: 1 # a at $sp+8, p at $sp+12 2 lw $t0, 8($sp) 3 lw $t1, 12($sp) 4 beq $t1, Szero, L3 5 ble $t0, Szero, L3 6 lw $t2, 0($t1) 7 add $t2, $t0, $t2 8 sw $t2, 0($t1) 9 L3: A. Write a goto version in C that performs the same computation and mimics the control flow of the assembly code. B. Explain why the assembly code contains two conditional branches, even though the C code has only one if statement.

Problem 7.4 When given the C code 1 void cond(int a, int p) 2 { 3 4 if (p && a > 0) p += a; a MIPS compiler generates the following assembly code for the body of the instruction: 1 # a at $sp+8, p at $sp+12 2 lw $t0, 8($sp) 3 lw $t1, 12($sp) 4 beq $t1, Szero, L3 5 ble $t0, Szero, L3 6 lw $t2, 0($t1) 7 add $t2, $t0, $t2 8 sw $t2, 0($t1) 9 L3: A. Write a goto version in C that performs the same computation and mimics the control flow of the assembly code. B. Explain why the assembly code contains two conditional branches, even though the C code has only one if statement.

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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// Example 8.27a void Func(int a[], int & r) { int i; for (i = 0; i < 100; i++) { a[i] = r + i/2; }} The Intel assembler produces the following assembly code from Example 8.27a (32 bit mode):
[2] ( Explain for each line why you chose a certain suffix such as b,w,l or q For each of the following lines of assembly language, determine the appropriate instruction suffix based on the operands. (For example, mov can be rewritten as movb, movw, movl, or movq.) mov mov mov που mov που Zeax, (%rap) (%rax), %dx $0xFF, %bl (%rap,%rdx, 4), %d1 (%rdx), %rax %dx, (%rax)
3. Find out the MIPS instructions for the following code: A = B - X[4]; If(A >= C){ C= 800; D = 50000; // 50000 = 0XC350 } elsel D = C+ 7000000; / 7000000 = 0X6ACFCO } X[8] = D - 400; %3D Assume the address of first instruction is 2000 in decimal. The program/saved registers and temporary registers are available as $s0 -- $s7 and $t0 - $t9 respectively.
[2] ( Explain for each line why you chose a certain suffix such as b,w,l or q For each of the following lines of assembly language, determine the appropriate instruction suffix based on the operands. (For example, mov can be rewritten as movb, movw, movl, or movq.) mov mov mov που mov mov %oax, (%rsp) (%rax), %dx $0xFF, %bl (%rap,%rdx, 4), %d1 %rdx), %rax %dx, (%rax)
1. A. Convert the following C code into AT&T assembly: x at 8(%ebp) if (x>5){ X++; } else { X--; } while (x<10){ x++; } B. Annotate each line of the assembly code to describe how it operates. C. Write a goto version of the function (in C) that mimics how the assembly code program operates.
[2] ( Explain for each line why you chose a certain suffix such as b,w,l or q For each of the following lines of assembly language, determine the appropriate instruction suffix based on the operands. (For example, mov can be rewritten as movb, movw, movl, or movq.) mov mov τον που mov mov Zeax, (%rap) (%rax), %dx $0xFF, %bl (%rap,%rdx, 4), %d1 %rdx), %rax %dx, (%rax)
15. By MIPS convention, $sp is the stack point register holding the value of the ending memory position for the stack, and Sao is the (5) register for storing the first parameter of a call to a function named X. De-compile the following assemble codes back to the c/C++/Java function X. X: addi $sp, $sp, -8 $ra, 4 ($sp) $a0, o($sp) Sw Sw $t0, $a0,1 $t0, $zero, L1 slti # test for $a0 < 1, if not, $to gets 1, else $to gets o beg addi $vo, $zero,0 $sp, $sp,8 $ra addi jr L1: addi Şa0, $a0,-1 jal x $a0, 0($sp) $ra, 4($sp) addi $sp, $sp, 8 add $vo, Şao, $vo jr $ra lw lw
mystery: mov %rdi, %rax cmp %rsi, %rax cmovl %rsi, %rax cmp %rdx, %rax cmovl %rdx, %rax ret Parameters x, Y, z are stored in registers %rdi, %rsi, and %rdx, respectively. Write C code for mystery that will have an effect equivalent to the assembly code.
The instruction that is not possible among the following is a) MOV AX, BX b) MOV AX, [BX]. c) MOV 55H, BL d) MOV AL, 55H
2. [4pts] Use the following C-Code for the problems below. int recFunc (int a, int b) { if (b = 0) == return a; else return 1+recFunc (a, b-1); a. Give the flowchart for the C-Code b. Convert to MIPS assembly and comment each assembly instruction to indicate corresponding C-Code.
Systems Programming Purpose: The purpose of this assignment is to practice loop instructions and array iterations in M6800 assembly language programming. Write an assembly language program which implements the following C code that is a find minimum problem solution implementation: int A[10] = {8, 14, 6, 16, 5, 5, 10, 9, 4, 11}; int i = 0; int min = 255; // This is the largest number 8-bit accumulators can hold. while (i < 10) { if (A[i] < min) min = A[i]; } Your solution should be able to handle all possible array elements and orderings for any value an accumulator can hold. You should treat variable į as XR in assembler (index register) and store variable min in the address 60H. 0010: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00 8628: 00 060 00 00 00 600 00 00 00 08 08 00 6 00 00 00 0030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0050: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 006 0: 00 00 00 00 00 00 00 00 00 08 00…
Description In this lab task you will translate and implement an if -else pseudocode decision into assembly language. Implement the following pseudocode in x86 assembly language. Use short-circuit evaluation and assume that num1 and RESULT are 32-bit variables. if ( num1 > ecx ) AND ( ecx > edx ) RESULT = 1 else RESULT = 2; Title your branch label using the 1st letter of your first name and last name. For example, Ann Napa's conditional branch label would be AN: Step through your program with a debugger to see exactly what is going on as you test your program branching execution.