Find odd vector elements C++ - c++

Question

So I have written a small program that calculates A[0]*B[0] - A[1]*B[1] - A[2]*B[2] - ... - A[n-1]*B[n-1] using vectors.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
double CalcProduct(vector<double> v1,vector<double> v2){
size_t size; int i = 0;
vector<double> product;
double input, input2;
cout << "Enter vector size: ";
cin >> size;
cout << "Enter values for first vector: ";
while (i++ != size) {
cin >> input;
v1.push_back(input);
}
i = 0;
cout << "Enter values for second vector: ";
while (i++ != size) {
cin >> input2;
v2.push_back(input2);
}
double result = 0.0;
if (size >= 1) {
result += v1[0] * v2[0];
for (int i = 1; i < size; ++i)
result -= v1[i] * v2[i];
}
cout << "\nThe 'happy' product of both vectors is: " << result << endl;
}
int main()
{
vector<double> v1;
vector<double> v2;
CalcProduct(v1,v2);
return 0;
}
However, what would happen if I want to compute A[1]*B[1] - A[3]*B[3] - A[5]*B[5] - ....? I tried adding +1 to v1 and v2 before they sum up in result but my logic seems to be wrong. Sample input:
5
1 2 3 4 5
4 5 6 7 5
Expected output: -18

Answer 1

You're doing the calculation in this loop
for(int i = 1; i < size; ++i)
result -= v1[i] * v2[i];
so instead of incrementing i, you can make jumps of 2:
for(int i = 1; i < size; i+=2) {
result -= v1[i] * v2[i];
}
Note that it's a good practice to check that you're not running out of bounds, also I like to surround body of for loop with { and } even if it contains one line.
Edit: I didn't note that you already added the first element before the loop, you can easily solve this by starting from i=3 instead of 1.

Answer 2

First of all, calcProduct is doing three things at once: Reading the values, calculating the product and writing the result to the output. Hence, you should split that into two methods and maybe do the printout from main().
Secondly, once you have generated the vectors, you can use vector.size() to avoid out of bounds situations by checking loop counters against it.
// give vector as reference
void ReadInput(vector<double>& v1,vector<double>& v2)
{
size_t size = 0; // use size_t here as this will be well defined
size_t i = 0; // on 32 and 64 bit architectures
// maybe you should have more error handling here as
// one might desire size to be in some application
// specific range
cout << "Enter vector size: ";
cin >> size;
cout << "Enter values for first vector: ";
while (i++ != size)
{
cin >> input;
v1.push_back(input);
}
i = 0;
cout << "Enter values for second vector: ";
while (i++ != size)
{
cin >> input2;
v2.push_back(input2);
}
}
// giving the parameters as (const) references avoids unneccessary copying
// of data
double CalcProduct(const vector<double>& v1, const vector<double>& v2,
size_t start, size_t step)
{
double result = 0.0;
// do parameter checking before the calculation in order to keep
// the code more structured (e.g. this can be move to a seperate
// function if it gets too complex
if ( v1.size() != v2.size()
|| start > v1.size()
|| start+step > v1.size())
|| start < 1 )
{
throw out_of_range("CalcProduc(): parameter mismatch");
}
result = v1[start] * v2[start];
for (size_t i = start+step; i < v1.size(); i+=step)
{
result -= v1[i] * v2[i];
}
return result;
}
int main()
{
vector<double> v1;
vector<double> v2;
ReadInput(v1, v2);
cout << "\nThe 'happy' product of both vectors is: " << CalcProduct(v1, v2, 1, 2) << endl;
return 0;
}