mp-units/latest/blog/2026/07/24/introducing-logarithmic-quantities-and-units/

[giscus[bot]]

mp-units/latest/blog/2026/07/24/introducing-logarithmic-quantities-and-units/

The quantities and units library for C++

https://mpusz.github.io/mp-units/latest/blog/2026/07/24/introducing-logarithmic-quantities-and-units/

[doocman]

I would probably have to re-read this a couple of times to really internalise the design decisions discussed in this blogpost. I have a comment from a statistics/information theory point of view. In computational statistics, the notion of log-likelyhood is quite common: it may be quicker to calculate (when dealing with gaussian distributions at least) and the regular likelyhood may very quickly truncate to zero. Another reason to work with the log-domain is that we can in some places replace multiplications and divisions with additions and subtractions.
One example: the PDF of a one-dimensional Gaussian random variable is 1/sqrt(2 pi sigma^2) * exp(-1/2 * (x - mean)^2/sigma^2), turn it into logarithm domain gives us: -1/2(log(2 * pi) + 2 log(sigma) + (x - mean)^2/sigma^2). Note the units here: log(2 pi) is unitless (and may be omitted if we are e.g. comparing different gaussian distributions against each others), 2 log(sigma) is log(<unit of variable^2>) and the final term is unitless. But the expression has unit log(1/<unit of variable>) or -log(<unit of variable>). This type of expression can be useful when an 'entropy weighted' comparison is needed. (log here refers to the natural logarithm).

I also wonder about the section with 'sqrt(): I would personally be surprised if it returns /2. I would expect that operations on a log-quantity also operates in the log-domain. One idea could be to have a linear_sqrt`-function that makes it clear in code that it operates as a square-root in the linear domain, and behaves differently in other domains.

[mpusz]

Thank you, Robin. The article is served under two URLs, and giscus opened this thread separately from #827, where the earlier comments are. I copied your comment there with attribution and answered it, so please continue in #827.