Question 3 We use a 32-bit operating system. Its int types, memory addresses, and registers, such as ESP, EBP and EIP, are represented by 32 bits. We compile a program that calls this function in this operating system. void foo (int **input, int par) { int tag = par; int *arr[10]; printf("%x", tag); if (tag > 1) { while (*input != 0x0000) { *arr = *input; arr++; input++; } } printf(“%x", tag); } (1) Please plot the stack frame of foo function before it returns.
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Please plot the stack frame of foo function before it returns.
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- int i; float x[5], y[5]; [ lots of code deleted... ] for (i=0;i y[i]) x[i] = x[i] + y[i]; [ more code deleted... ] Translate to MIPS assembly language:12.14 Zylab 3 - Single Procedure Call Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the original array raised to the index, i.e., P[i] = A[i]^i. For this, write two functions that will be called in main function independently. ● power inputs: element (A[i]) and index (1) • task: returns the value of element raised to index (A[i]^i). } O • newElement inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i) o task: add the new element at the end. o This function does not return any value (void). O Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be changed. int main() { // Variable Declaration int *A, *P; int n, k; int pow; // Task of main function P[0] 1; for (int j = k = j; pow // Base addresses of A and P // Lengths of arrays A and B // Return value from power function }…Consider the following C code: int A[16]; int B[16]; int m; ... //A large chunk of code that does NOT access //arrays A and B. .. for (int i=0; i<10; i++) { for (int j=0; j<16; j++) { B[j] = m * A[j] + B[j]; } } Assume this program runs on a 32-bit machine, i.e., the CPU loads/stores 4 bytes from memory in one go. This machine has a 16-bit memory address, and each memory block stores 16 bytes. This machine has a direct-mapped data cache with 16 cache lines. Array A starts at address O, and B starts at address 256 - both arrays begin at a memory block boundary. Each element of arrays A and B occupies 4 bytes. The values of i, j, and m are stored in CPU registers. Q5.1 A compulsory cache miss happens the first time the CPU reads any bytes in a memory block. Such cache misses are inevitable, and the entire memory block must be brought into the cache. How many compulsory misses in the data cache will occur when running the above code? Give your answer in a base 10 number (and only write…
- CODE IN C++ PLEASE....4. The Single Instruction Computer (SIC) has only one instruction that can do all operations our RISC-V does (you did a homework problem). The instruction has the following format sbn a, b, c # Mem[a]=Mem[a]- Mem[b]; if (Mem[a]- Mem[b]<0) go to PC+c For example, here is the program to copy a number from location a to location b: sbn temp, temp, 1 sbn temp, a, 1 sbn b,b, 1 sbn b, temp 1 Start: Design a single cycle datapath and control for this instruction set architecture.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…
- Computer Science #include<cmath>#include<stdio.h>__global__voidprocess_kernel1(float *input1,float *input2,float *output,int datasize){int idx = threadIdx.x + blockIdx.x * blockDim.x;int idy = threadIdx.y + blockIdx.y * blockDim.y;int idz = threadIdx.z + blockIdx.z * blockDim.z;int index = idz * (gridDim.x * blockDim.x) * (gridDim.y*blockDim.y) + idy * (gridDim.x * blockDim.x) +idx;if(index<datasize)output[index] = sinf(input1[index]) + cosf(input2[index]);}__global__voidprocess_kernel2(float *input,float *output,int datasize){int idx = threadIdx.x + blockIdx.x * blockDim.x;int idy = threadIdx.y + blockIdx.y * blockDim.y;int idz = threadIdx.z + blockIdx.z * blockDim.z;int index = idz * (gridDim.x * blockDim.x) * (gridDim.y*blockDim.y) + idy * (gridDim.x * blockDim.x) +idx;if(index<datasize)output[index] = logf(input[index]);}_global__voidprocess_kernel3(float *input,float *output,int datasize){int idx = threadIdx.x + blockIdx.x *…Software: sim8085 Write 8085 microprocessor code and attach output screenshot.Write a C language function named count that takes a void pointer and int parameter, representing a memory buffer, and a uint8_t parameter representing a byte, and returns an int result. The returned result should be the number of occurrences of the byte in the bufer. For example if the buffer p pointed to the bytes 01 12 01 23, then count(p, 4, 0x01) should return 2.
- convert this code to java language /* OPERATING SYSTEMS LAB PROJECT* AKASH JAIN* 179303013* DESIGNING A VIRTUAL MEMORY MANAGER*/ #include<stdio.h>#include<stdlib.h>#include<string.h> const int VM_SIZE=256;const int PAGE_SIZE=256;const int TLB_SIZE=16;const int MM_SIZE=256; int main(int argc, char* argv[]){FILE *fd;if(argc<2){printf("NOT ENOUGH AGRUMENTS\n EXITING\n");return 0;}fd=fopen(argv[1],"r");if(fd==NULL){printf("ERROR OPENING FILE\n FILE FAILED TO OPEN\n");return 0;}char *value=NULL;size_t len=0;ssize_t read;long long page_no,offset,page_table,totalhits=0,fault=0,pages=0;int qp=0; //to maintain the queue positionint physicalad=0,frame,logicalad;int tlb[TLB_SIZE][2];int pagetable[PAGE_SIZE]; memset(tlb,-1,TLB_SIZE*2*sizeof(tlb[0][0]));memset(pagetable,-1,sizeof(pagetable));int mask=255,mask1=62580,i,hit;while((read=getline(&value,&len,fd))!=-1){pages++;//get page number and offset from logical…int i = 5; is a statement in a C program. A. during execution, value of i may change but not its address B. during execution both the address and value may change C. repeated execution may result in different addresses for i D. i may not have an associated addressTranslate C program to Pep/9 assembly language. /C code for 6.19a#include <stdio.h>char myChar;char toLower(char ch) { if ((ch >= ’A’) && (ch <= ’Z’)) { return ch + ’a’ - ’A’; } else { return ch; }}int main () { scanf(”%c”, &myChar); printf(”%c\n”, toLower(myChar)); return 0;}