Write an assembly language program that corresponds to the following C program:     const int amount = 20000; int num; int sum;   int main () {    scanf("%d", &num);    sum = num + amount;    printf("sum = %d\n", sum);    return 0; }

Example: The Problem Input File Using C programming language write a program that simulates a variant of the Tiny Machine Architecture. In this implementation memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the Tiny Machine Architecture: //IN 5 //OUT 7 //STORE O //IN 5 //OUT 7 //STORE 1 //LOAD O //SUB 1 55 67 30 55 67 1 LOAD 2- ADD 3> STORE 4> SUB 5> IN 6> OUT 7> END 8> JMP 9> SKIPZ 31 10 41 30 //STORE O 67 //OUT 7 11 /LOAD 1 //OUT 7 //END 67 70 Output Specifications Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program. Your simulator should provide output according to the input file. Along with…

Using the following C code and assembly code answer the following;  C: long int nothingGreater(long int num) { if (num > 240) { return 240 - num; } else { return num; } } Assembly:  0x00000000004004ed <+0>: push %rbp 0x00000000004004ee <+1>: mov %rsp,%rbp 0x00000000004004f1 <+4>: mov %rdi,-0x8(%rbp) 0x00000000004004f5 <+8>: cmpq $0xf0,-0x8(%rbp) 0x00000000004004fd <+16>: jle 0x40050a <nothingGreater+29> 0x00000000004004ff <+18>: mov $0xf0,%eax 0x0000000000400504 <+23>: sub -0x8(%rbp),%rax 0x0000000000400508 <+27>: jmp 0x40050e <nothingGreater+33> 0x000000000040050a <+29>: mov -0x8(%rbp),%rax 0x000000000040050e <+33>: pop %rbp 0x000000000040050f <+34>: retq  What are the memory addresses of the three statements that implement the if statement (provide hex addresses with 0x prefix and without leading zeroes)? The first two statements that work together: The third statement that skips the else code if the if…

Convert this following C++ code below to easy68k assembly language: int sum = 0; int c =1; for (unsigned short i = 0; i <= 10; i++) { if (c == 1 ) { sum += i; c = 0; } else { c = 1; } }