How to multiply a vector and scalar in C++? - c++

Question

I would like to multiply a vector with a scalar. This vector was created using the accepted answer to this question of mine namely:
std::vector<int> n(N + 1);
std::iota(begin(n), end(n), 0);
and I would like to multiply this vector, n, with a scalar (specifically of type double, if it is relevant here) called npi.
I have seen this answer to a previous question here, but it wasn't all that helpful. The way I attempted to implement it was by adding:
std::transform(n.begin(), n.end(), n.begin(),
std::bind1st(std::multiplies<T>(),pin));
to my C++ program. This returned the compile error:
error: ‘T’ was not declared in this scope
std::bind1st(std::multiplies<T>(),pin));
I would like to call the vector created by multiplying this vector with a scalar npi, so please do not give me code that will call this new vector n (i.e., overwriting my existing n vector).
EDIT:
If it will placate whomever voted to close this question, here is my full program:
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
#include <cmath>
#include <utility>
#include <unistd.h>
#include <algorithm>
#include <numeric>
/*#include <armadillo>*/
using namespace std;
/*using namespace arma;*/
double N = 1000.0;
double x0 = 0;
double x1 = 100;
double pin = M_PI / double(N);
int main() {
std::vector<int> n(N + 1);
std::iota(begin(n), end(n), 0);
std::transform(n.begin(), n.end(), n.begin(),
std::bind1st(std::multiplies<T>(),pin));
for(double i: n)
{
std::cout << i << '\n' << std::scientific;
}
}

Answer 1

For vector<int> output, one way is:
auto npi = n;
for( auto& i: npi )
i *= pin;
If npi should be vector<double> (not clear from the question) then replace the first line with:
std::vector<double> npi( n.begin(), n.end() );

Answer 2

You need to replace T by the type contained in the vector, in this case int. However you can probably simplify your code by using a lambda function here instead:
#include <algorithm> // for std::transform
#include <cmath> // for M_PI
#include <iostream> // for std::cout etc
#include <numeric> // for std::iota
#include <vector> // for awesome
int main() {
std::vector<int> vec1(10);
std::iota(vec1.begin(), vec1.end(), 0);
int N = 42;
std::vector<double> vec2(vec1.size()); // vec2 needs to be as big or bigger than vec1
std::transform(vec1.begin(), vec1.end(), vec2.begin(),
[N](int i) { return i * M_PI / N; });
for (auto a : vec1)
std::cout << a << " ";
std::cout << std::endl;
for (auto a : vec2)
std::cout << a << " ";
std::cout << std::endl;
}
Here's an online example: http://melpon.org/wandbox/permlink/XrNxDND0steJmym8

Answer 3

If I have understood you correctly you need the following
std::vector<double> v;
v.reserve(n.size());
std::transform(n.begin(), n.end(), std::back_inserter( v ),
std::bind1st(std::multiplies<double>(), pin));

Answer 4

You can pass the scalar in the capture clause of the Lambda function and do the multiplication inside the lambda function itself
#include <algorithm>
#include <vector>
std::vector<int> foo;
std::vector<int> bar;
auto npi=4.0;
std::transform (foo.begin(), foo.end(), bar.begin(), foo.begin(), [&npi](auto& c){return c * npi;}