Computing the scalar product of two vectors in C++ - c++

Question

I am trying to write a program with a function double_product(vector<double> a, vector<double> b) that computes the scalar product of two vectors. The scalar product is
$a_{0}b_{0}+a_{1}b_{1}+...+a_{n-1}b_{n-1}$.
Here is what I have. It is a mess, but I am trying!
#include <iostream>
#include <vector>
using namespace std;
class Scalar_product
{
public:
Scalar_product(vector<double> a, vector<double> b);
};
double scalar_product(vector<double> a, vector<double> b)
{
double product = 0;
for (int i = 0; i <= a.size()-1; i++)
for (int i = 0; i <= b.size()-1; i++)
product = product + (a[i])*(b[i]);
return product;
}
int main() {
cout << product << endl;
return 0;
}

Answer 1

Unless you need to do this on your own (e.g., writing it is homework), you should really use the standard algorithm that's already written to do exactly what you want:
#include <iostream>
#include <numeric>
#include <vector>
int main() {
std::vector<double> a {1, 2, 3};
std::vector<double> b {4, 5, 6};
std::cout << "The scalar product is: "
<< std::inner_product(std::begin(a), std::end(a), std::begin(b), 0.0);
return 0;
}
Note that while begin(a) and end(a) are new in C++11, std::inner_product has been available since C++98. If you are using C++ 98 (or 03), it's pretty easy to write your own equivalent of begin and end to work with arrays though:
template <class T, size_t N>
T *begin(T (&array)[N]) {
return array;
}
template <class T, size_t N>
T *end(T (&array)[N]) {
return array + N;
}
Using these, a C++ 98 version of the previous code could look something like this:
int main() {
double a[] = {1, 2, 3};
double b[] = {4, 5, 6};
std::cout << "The scalar product is: "
<< std::inner_product(begin(a), end(a), begin(b), 0.0);
return 0;
}
Note that the begin and end above will only work for arrays, where the begin and end in C++11 (and later) will also work for normal collection types that define a .begin() and .end() (though it's trivial to add overloads to handle those as well, of course):
template <class Coll>
typename Coll::iterator begin(Coll const& c) { return c.begin(); }
template <class Coll>
typename Coll::iterator end(Coll const& c) { return c.end(); }

Answer 2

You can delete the class you have defined. You don't need it.
In your scalar_product function:
double scalar_product(vector<double> a, vector<double> b)
{
double product = 0;
for (int i = 0; i <= a.size()-1; i++)
for (int i = 0; i <= b.size()-1; i++)
product = product + (a[i])*(b[i]);
return product;
}
It's almost there. You don't need 2 loops. Just one.
double scalar_product(vector<double> a, vector<double> b)
{
if( a.size() != b.size() ) // error check
{
puts( "Error a's size not equal to b's size" ) ;
return -1 ; // not defined
}
// compute
double product = 0;
for (int i = 0; i <= a.size()-1; i++)
product += (a[i])*(b[i]); // += means add to product
return product;
}
Now to call this function, you need to create 2 vector objects in your main(), fill them with values, (the same number of values of course!) and then call scalar_product( first_vector_that_you_create, second_vector_object );

Answer 3

While you have been presented many solutions that work, let me spin up another variation to introduce a couple of concepts that should help you writing better code:
class are only needed to pack data together
a function should check its preconditions as soon as possible, those should be documented
a function should have postconditions, those should be documented
code reuse is the cornerstone of maintenable programs
With that in mind:
// Takes two vectors of the same size and computes their scalar product
// Returns a positive value
double scalar_product(std::vector<double> const& a, std::vector<double> const& b)
{
if (a.size() != b.size()) { throw std::runtime_error("different sizes"); }
return std::inner_product(a.begin(), a.end(), b.begin(), 0.0);
} // scalar_product
You could decide to use the inner_product algorithm directly but let's face it:
it requires four arguments, not two
it does not check for its arguments being of the same size
so it's better to wrap it.
Note: I used const& to indicate to the compiler not to copy the vectors.

Answer 4

You seem to want to make a class specifically for vectors. The class I made in my example is tailored to 3 dimensional vectors, but you can change it to another if desired. The class holds i,j,k but also can conduct a scalar products based on other MathVectors. The other vector is passed in via a C++ reference. It is hard to deduce what the question was, but I think this might answer it.
#include <iostream>
using namespace std;
class MathVector
{
private:
double i,j,k;
public:
MathVector(double i,double j,double k)
{
this->i=i;
this->j=j;
this->k=k;
}
double getI(){return i;}
double getJ(){return j;}
double getK(){return k;}
double scalar(MathVector &other)
{
return (i*other.getI())+(j*other.getJ())+(k*other.getK());
}
};
int main(int argc, char **argv)
{
MathVector a(1,2,5), b(2,4,1);
cout << a.scalar(b) << endl;
return 0;
}

Answer 5

Here is the code that you should have. I see you have used class in your code, which you do not really need here. Let me know if the question required you to use class.
As you are new and this code might scare you. So, I will try to explain this as I go. Look for comments in the code to understand what is being done and ask if you do not understand.
//Scalar.cpp
#include <stdlib.h>
#include <iostream>
#include <vector>
using namespace std;
/**
This function returns the scalar product of two vectors "a" and "b"
*/
double scalar_product(vector<double> a, vector<double> b)
{
//In C++, you should declare every variable before you use it. So, you declare product and initialize it to 0.
double product = 0;
//Here you check whether the two vectors are of equal size. If they are not then the vectors cannot be multiplied for scalar product.
if(a.size()!=b.size()){
cout << "Vectors are not of the same size and hence the scalar product cannot be calculated" << endl;
return -1; //Note: This -1 is not the answer, but just a number indicating that the product is not possible. Some pair of vectors might actually have a -1, but in that case you will not see the error above.
}
//you loop through the vectors. As bobo also pointed you do not need two loops.
for (int i = 0; i < a.size(); i++)
{
product = product + a[i]*b[i];
}
//finally you return the product
return product;
}
//This is your main function that will be executed before anything else.
int main() {
//you declare two vectors "veca" and "vecb" of length 2 each
vector<double> veca(2);
vector<double> vecb(2);
//put some random values into the vectors
veca[0] = 1.5;
veca[1] = .7;
vecb[0] = 1.0;
vecb[1] = .7;
//This is important! You called the function you just defined above with the two parameters as "veca" and "vecb". I hope this cout is simple!
cout << scalar_product(veca,vecb) << endl;
}
If you are using an IDE then just compile and run. If you are using command-line on a Unix-based system with g++ compiler, this is what you will do (where Scalar.cpp is the file containing code):
g++ Scalar.cpp -o scalar
To run it simply type
./scalar
You should get 1.99 as the output of the above program.