I am trying to understand the dimensionality of the following two equations.

--- first equation ---
example::types::radii_t val1 = 5 * example::types::radii;
example::types::radii_t val2 = 4 * example::types::radii;

example::types::radii_t L1 = 1 / ( val1 - va2 );

va2->val2

If I was doing the dimensional analysis by hand I would see

1 / ( radii - radii )

Such that the end type would be

radii^-1

Given that end type here is the second equation:

example::types::radii_t val3 = 5 * example::types::radii;
example::types::dimensionless_t val4 = 10;
example::types::dimensionless_t L2 = L1 * val4 * val3;

Doing the dimensional analysis by hand I think I should see:

L2 = radii * (nothing) * radii^-1
= (nothing) or dimensionless

What Boost Units is reporting is that L2 is actually units in radii_t.

I'm not sure I believe you; the code you pasted is not compilable...did you actually compile anything?

--------- QUESTIONS --------------

Q1: How do you print out the types at compile time?

Q2: How can you confirm you have the correc power for a type? For
example radii is indeed raised to the -1 power.

#include "types.hpp"

#include <iostream>

#include <boost/units/io.hpp>

using namespace myproject::types;

int main()

{

radii_t val1(5*radii),

val2(4*radii);

//auto L1 = 1/(val1-val2); // compile time error due to failed dimensional consistency

radii_t val3(5*radii);

dimensionless_t val4 = 10;

dimensionless_t L2(val3*val4/(val1-val2));

std::cout << val1 << std::endl

<< val2 << std::endl

<< val3 << std::endl

<< val4 << std::endl

<< L2 << std::endl;

return 0;

}