Answered: For the following C statement, write a…

For the following C statement, write a minimal sequence of MIPS assembly instructions that does the identical operation. int arr[] - {0x11,0x22, ®x33, ®x44, ®x55}; arr[2] = arr[0] + 3;

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

See similar textbooks

Similar questions

Fill in the blanks to match with the assembly code.
Convert the following C++ programs into Pep/9 assembly #include <iostrea> using namespace std; int minimum (int i1, int i2){if (i1 < i2)return i1;elsereturn i2;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
The following instruction set is supported by a simple processor, which is similar to what we discussed in the class, with a few new instructions added. The format of most instructions is defined as follows. bits 15:14 13:10 9 8:6 5:3 2:0 field unused opcode w srcl src2 dst where the fields are defined as follows. opcode : operation to be performed by the processor write back ALU output to register file (1= yes, 0 = no) address of the first ALU operand in the register file address of the second ALU operand in the register file address in the register file where the output is written w: srcl: src2: dst: For opcodes BEQ, BLEZ and JUMP, the 6 least significant bits (5:0) give an address in the instruction memory, which is byte-addressed. The opcode HALT has all operand bits (9:0) being 0. When an instruction has only two operands, the field for the unused operand is filled with 0-bits. For example, bits (5:3) for SLL are all zero because src2 is not used. The opcode and meaning of these…
Convert the following C++ programs into Pep/9 assembly #include <iostrea> using namespace std; int minimum (int i1, int i2){if (i1 < i2)return i1;elsereturn i2;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
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 The instruction mov $0xf0,%eax at address 0x4004ff puts a value into register %eax but there are no other references to %eax in the program. Which instruction makes use of this value? What is the address of the instruction (0x-prefixed hex)? What…
Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; int minimum (int i1, int i2){if (i1 < i2)return i1;elsereturn i2;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Please change this C code to Assembly program(x86 and ARM assembly gcc) #include <stdio.h>int main() { unsigned guess; /* current guess for prime */ unsigned factor; /* possible factor of guess */ unsigned limit; /* find primes up to this value */ printf("Find primes up to: "); scanf("%u", &limit); printf("2\n"); /* treat first two primes as special case */ printf("3\n"); guess = 5; /* initial guess */ while ( guess <= limit ) { /* look for a factor of guess */ factor = 3; while ( factor*factor < guess && guess % factor != 0 ) factor += 2;if ( guess % factor != 0 ) printf("%d\n", guess); guess += 2; /* only look at odd numbers */ }printf("Recieves a number(limit) and prints its all divisors"); for(int i = 1; i <= limit; i++) { if((limit%i) == 0){ printf("\n%d", i); } } return 0; }
You may use only the instructions we covered in class: add, sub, addi, lw, sw, and the same rules as mentioned earlier still apply.AssignmentWrite MIPS assembly code implementing the following C/C++ statement:a[k*2-4] = a[k]*2 + 7;
Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; int minimum (int i1, int i2){if (i1 < i2)return i1;elsereturn i2;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Convert the following C++ program into Pep/9 Assembly. #include <iostream>using namespace std;void SetEqualGreater (int &num1, int &num2){ if (num1 < num2) num1 = num2; else num2 = num1;}int main() { int x, y; // Using Local Variables cin >> x >> y; SetEqualGreater(x, y); cout << "x = y = " << x << endl; return 0;} Submit: Source file along with screen captures showing the program running in the Pep/8 simulator.
Consider the following C code: 1: int main() { 2: 3: 4: 5: } int y = 5; // Return y times 4 return y* 4; What is removed during the preprocessing step of compilation? A. Line 1 B. Line 3 C. Line 5 D. Lines 3 and 5 E. Everything, since it will be assembly after the preprocessing step
Select common examples of when an assembly programmer would want to use the stack: to pass arguments to save return address for CALL local variables temporary save area for registers applications which have FIFO nature, such as customers waiting in a bank queue