Re: [Boost-users] Problem with user defined Boost Units system

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 This should give you a compile error anyway, since you are trying to assign a length to an inverse length...

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.

Using your earlier header : #include "types.hpp" #include <iostream> #include 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; } gives 5 r 4 r 5 r 10 dimensionless 50 dimensionless