Compute the square root of the sum of the squares of an array in C++

By: Ignatius

This simple C++ program computes the square root of the sum of the squares of all the positive elements in array named x. It uses the header file math.h since it contains the mathematical functions pow() (for taking the power of a number) and sqrt() (for taking the square root of a number)

#include <iostream> using namespace std; #include <cmath> #define m 4 #define n 5 int main() { int i, j; int x[m][n]={{4,5,6,2,12},{10,25,33,22,11}, {21,32,43,54,65},{3,2,1,5,6}}; float sum2, result; // the outer for loop, read row by row... for(i=0; i<m; i++) { // the inner for loop, for every row, read column by column for(j=0; j<n; j++) { // set some condition here to avoid divides by 0... if(x[i][j]>0) // do the square of the array elements and then sum up... sum2 = sum2 + pow(x[i][j], 2); } // assign the result to variable result, do the square root on the previous result.... result = sqrt(sum2); } // some story and printing the result... cout<<"\nFirst, summing up all the arrays' element"; cout<<"\nThe given array has 4 x 5 in size,\n"; cout<<"\nThe sum is = "<<sum2; cout<<"\nNext, square root the sum\n"; cout<<"\nThe answer is = "<<result<<"\n"; return 0 }