Given the following code segment in C++, please write its corresponding EASy68K assembly code. Please write code in EASY68K assembly code
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Given the following code segment in C++, please write its corresponding EASy68K assembly code.
Please write code in EASY68K assembly code!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
int sum = 0
; int c =1;
for (unsigned short i = 0; i <= 10; i++)
{ if (c == 1 ) {
sum += i;
c = 0;
}
else {
c = 1;
}
}
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Solved in 2 steps
- Can u show me the out put of this assembly language with different in put please INCLUDE Irvine32.inc.stack 4096ExitProcess proto,dwExitCode:dword.data; strings(texts) for promptsprompt_1st_int BYTE "Please enter the first integer: ",0prompt_2nd_int BYTE "Please enter the second integer: ",0prompt_text_color BYTE "Please pick the text color (1=blue, 2=green, 4=red, or any number from 0-15): ",0prompt_proverb BYTE "Please teach me a proverb: ",0response_str BYTE "The proverb I just learned: ", 0; strings(texts) for output labels/descriptionsdec_sum_str BYTE "The sum in DEC is: ",0hex_sum_str BYTE "The sum in HEX is: ",0memory_dump_str BYTE "Memory Dump with operand_1, _2, and sum: ",0; variables definitionsoperand_1 SDWORD ?operand_2 SDWORD ?sum SDWORD ?color_code DWORD ?aString BYTE 80 DUP(0), 0.codemain proc;prompt the 1st number inputmov edx, OFFSET prompt_1st_intcall WriteString; set input text color, "yellow" is a pre-define constant, in the Irvine librarymov eax, yellowcall…Using signed 16-bit numbers, convert the following statement in C++ to Assembly language. (Example of an assembly language and how it should look like: movf i,w subwf j,w movf i+1, w subwfb j+1,w bc end_if movf j,w addwf i,w movwf m movf j+1,w addwfc i+1,w movwf m+1 end_if) signed char i, j, k; if (i < j) { k = j - i; } // … rest of code …Computer Science Write an x86 assembly program to count the odd numbers up to a NUMBER? Assume NUMBER variable holds decimal 10. So, RESULT should hold 5 (there are five odd numbers 1,3,5,7,9) (Your algorithm should also work for different values. You can assume RESULT is less than FFh) .MODEL SMALL.STACK 64.DATANUMBER DB 10RESULT DB ?.CODE
- Compile the following C code snippet with different processor architecture. int square(int num) { return num * num; } 1) compile using armv7-a clang 11.0, and explain the program. 2) compile using x86-64 gcc 11.2, and explain the program.In c++ write an assembler in which it will read a program written in HACK assembly language from an external file and ultimatley translate each line of code into the binary equivalent that can be run on the computer I built so based off the following hdl files Computer below others are in the images. CHIP Computer { IN reset; PARTS: //Read-only memory (ROM) for instruction fetch ROM32K(address=PC,out=instruction); //Central Processing Unit (CPU) for instruction executionCPU(instruction=instruction,reset=reset,inM=outMemo,outM=CPUoutM,writeM=wM,addressM=adM,pc=PC); //Memory for data storage and control logic Memory(in=CPUoutM,load=wM,address=adM,out=outMemo); }Write assembly code equivalent to the following C codeVoid myfunction(int *z , int x, int y ){int t = x;int b = y;*z= (t*b) +z +2;}
- Convert the given code fragment to assembly code fragment, using only instructions of the following types. These instructions are generally discussed in class. Here X,Y,Z are any memory locations; R, R1, R2 are any general registers; L is a label in the code (you can use any names as labels, ex. L, L1, L2 etc. ). load X, R //copy contents of memory location X into R. store R, X //Store contents of R into Mem location X cmp R1, R2 //Compute R1-R2 and update condition codes; //throw away result of subtraction. jmp L //Jump to location L in the code. jmpp L //If P bit is 1, Jump to location L in the code add X, R //Add contents of X,R and store result in R; //Also update the condition codes. Be careful about what type of argument is allowed in the instruction (Memory or Register). Ex. the first argument of ADD instruction is memory, not register. Do Not…turn this c code into assembly language in optimized and unoptimized mode. /* Project 2 */ /* GCC *//* gcc -Wa,-adhln -g -masm=intel -m32 "Project 2.c" > "Project 2-g.asm" *//* gcc -Wa,-adhln -O -masm=intel -m32 "Project 2.c" > "Project 2-o.asm" */ #include <stdio.h> #define NOINLINE __attribute__ ((noinline)) static NOINLINE int function1(int x, int y){ int i; int sum; int values[10]; sum = 0; for (i = 0; i < 10; i++) { values[i] = 10 * i + x * y; sum += values[i]; } return (sum); } static int NOINLINE function2(int *values, int valuesLen){ int i; int sum; sum = 0; for (i = 0; i < valuesLen; i++) { sum += values[i]; } return (sum); } static NOINLINE int function3(int x){ int y; y = x / 10; return (y); } static NOINLINE int function4(int a, int b, int c, int d){ int r; if (a > b) r = a; else if (a > c) r = 2 * a; else if (a > d)…Assume we are using an unknown assembly language, without finding out what assembly language it is, try to figure out what these codes are going to do by GUESSING, and express them in high-level language such as Java or C. Code group 1: MOV R1, 0 LABLE1 ………… ADD R1, 1 CMP R1, 5 JLE LABLE1 Code group 2: MOV R1, 1 LABEL2 ………………. CMP R1, 10 JGE LABEL2 Code group 3: JMP LABEL3 LABEL4 …………….. LABEL3 CMP R1, 1 JE LABEL4
- For the following C statement, what is the corresponding MIPS assembly code? Assume that the variables f, g, h, and i are given and could be considered 32-bit integers as declared in a C program. Use a minimal number of MIPS assembly instructions. f = g + (h − 5);Please translate this to assembly from C language. Comment what each line of the code in assembly is doing. Please do not reference disassembly tools or compilers. Simply translate this to ARM assembly. If you are unsure please pass on this question. Thank you void SVN_SEG(unsigned int n){ int i=0; uint32_t temp=0; SEG_CTL=1; for(i=0; i<4; i++){ temp |= (n&0xF)<<(i*8); n>>4; } temp |= 0x80808080; SEG_DATA = temp; }This is what I have so far I need help with the commenting on the assembly code I have included my C code that I commented and this assembly was generated from my C code so it can be of help My assembly code largest(int*, int): push rbp mov rbp, rsp mov QWORD PTR [rbp-24], rdi mov DWORD PTR [rbp-28], esi mov rax, QWORD PTR [rbp-24] mov eax, DWORD PTR [rax] mov DWORD PTR [rbp-4], eax mov DWORD PTR [rbp-8], 1 jmp .L2 .L4: mov eax, DWORD PTR [rbp-8] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-24] add rax, rdx mov eax, DWORD PTR [rax] cmp DWORD PTR [rbp-4], eax jge .L3 mov eax, DWORD PTR [rbp-8] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-24] add rax, rdx mov eax, DWORD PTR [rax] mov DWORD PTR [rbp-4],…
