Investigate which int overloads are still needed (exp, pow, ...)

[rok-cesnovar]

Description

My lab mate asked me about the reasoning behind having exp(int) defined in the Math library.

Example: inline double exp(int x) { return std::exp(x); } in https://github.com/stan-dev/math/blob/develop/stan/math/prim/fun/exp.hpp#L18

I could not find a reason so I am wondering myself also. Can anyone help me understand the reasoning behind this.

This came up while trying to add a custom function in stan math and using something like exp(3.3) inside the stan::math namespace was being truncated down to exp(3). This was due to a missing using std::exp; that would have been picked if he used cpplint.

Current Version:

v3.1.0

[bob-carpenter]

That problem hadn't occurred to me!

The reason we have the int signature is to avoid ambiguity with exp(int) when std::exp() and stan::math::exp() are in scope. None of them have integer signatures, so an int either needs to be promoted to double or to var or fvar.

We could get around this by having our math lib version be defined as something like:

template <typename T, typename = std::enable_if_t<std::is_arithmetic<scalar_type_t<T>>::value>>
inline T exp(T x) {
  return std::exp(x);
}

Then it won't try to cast double to int, while still being a better match for exp(int) than std::exp().

[bob-carpenter]

OK, I just went and looked up the signatures of exp and as of C++11, there's an exp(IntegralType). That must not cause ambiguities with our exp(int), but it probably means we can get rid of our exp(int) definition without problems.

[rok-cesnovar]

while still being a better match for exp(int) than std::exp()

But stan::math::exp still uses std::exp so these would be the same in the end right? So I guess this was to prevent int being promoted to var?

[bob-carpenter]

Right---the exp(int) signature was just to disambiguate calls to exp(int) in Stan, which were otherwise ambiguous betwene promotion to double and promotion to var.

…

On Feb 10, 2020, at 3:02 PM, Rok Češnovar ***@***.***> wrote: while still being a better match for exp(int) than std::exp() But stan::math::exp still uses std::exp so these would be the same in the end right? So I guess this was to prevent int being promoted to var? — You are receiving this because you commented. Reply to this email directly, view it on GitHub, or unsubscribe.