calculate number of memory bytes accessed by this program: void my_dgemv(int n, double* A, double* x, double* y) { for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { y[i] += A[i * n + j] * x[j]; } } }
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calculate number of memory bytes accessed by this
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- int X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the pic1、Counting Primitive Operations void STRAITMAXMIN(A,n,max,min) //Set the maximum value in A to max and the minimum value to min { int i,n max=min=A[1] for i=2 to n { if A[i]> max max=A[i]; if A[i]< min min=A[i]; } }// add.ll define void @add(i32* %ptr1, i32* %ptr2, i32* %val) {ret void} Fill add.ll function to do the following operation: void add(int *ptr1, int *ptr2, int *val) { *ptr1 += *val; *ptr2 += *val; } This is the full question. It is related to LLVM. If it's going to help there is one more code given which is: #include <stdio.h> void add(int *ptr1, int *ptr2, int *val); int main(int argc, char **argv) {FILE *f = fopen(argv[1], "r");int a, b, c;fscanf(f, "%d %d %d", &a, &b, &c);add(&a, &b, &c);printf("%d %d\n", a, b);fclose(f); return 0;}
- JAVA CODE PLEASE Functions with 1D Arrays Quiz by CodeChum Admin Instruction: Write a function that accepts two integers X and Y and prints the binary representation of the numbers starting from X to Y. Note: X would always be lesser than Y. Input 1. integer X 2. integer Y Output Enter·X:·5 Enter·Y:·10 101·110·111·1000·1001·1010JAVA CODE PLEASE Functions with 1D Arrays Practice I by CodeChum Admin Instruction: Write a program that has a function that accepts a character array and counts the number of vowel in character array. Output Number·of·vowels:·5Java Functions with 1D Array Write a Function that accepts two integers X and Y and prints the binary representation of the numbers starting from X to Y. Note: X would always be lesser than Y. Input 1. Integer X 2. Integer Y 3. Integer X 4. Integer Y Output: Enter X: 5 Enter Y: 10 101 110 111 1000 1001 1010
- CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;Assignment for Computer Architecture: N Factual by Recusion *please have comments in the code* You are to write a program in MIPS that computes N! using recursion. Remember N! is the product of all the numbers from 1 to N inclusive, that is 1 x 2 x 3 x (N – 1) x N. It is defined as 1 for N = 0 and is undefined for values less than 0. The programs first requests the user to input the value of N (display a prompt first so the user knows what to do). If the input value is less than 0, the program is to display “N! undefined for values less than 0” and then requests the user to input the value of N again. If the value input is non-negative, it is to compute N! using a recursive function, that is one that calls itself. You are to have your name, the assignment number, and a brief description of the program in comments at the top of your program. Since this is an assembly language program, I expect to see comments on almost every line of code in the program. Also make the…void show_byte(byte_pointer start, int len) { Q2 int i; for(i=0; i#include <stdio.h> struct Single { int num; }; void printSingle(int f) { int binaryNum[33]; int i = 0; while(f>0) { binaryNum[i] = f % 2; f = f/2; i++; } for (int j=i-1; j>= 0; j--) { printf("%d",binaryNum[j]); } } int main() { struct Single single; single.num = 33; printf("Number: %d\n",single.num); printSingle(single.num); return 0; }#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…SEE MORE QUESTIONS