## Stretching samples

January 31, 2017

So for an experiment I ended up needing conversions between 8 bits and 16 bits samples. To upscale an 8 bit sample to 16 bits, it is not enough to simply shift it by 8 bits (or multiply it by 256, same difference) because the largest value you get isn’t 65535 but merely 65280. Fortunately, stretching correctly from 8 bit to 16 bit isn’t too difficult, even quite straightforward.

## Whatever sums your floats

January 24, 2017

While flipping the pages of a “Win this interview” book—just being curious, not looking for a new job—I saw this seemingly simple question: how would you compute the sum of a series of floats contained in a array? The book proceeded with the simple, obvious answer. But… is it that obvious?

## Choosing the Right Pseudoinverse

January 17, 2017

On a number of previous occasions, I have used the pseudoinverse of a matrix solve systems of equations, and do other things such as channel mixing. However, the demonstration I gave before isn’t entirely correct. Let’s see now why it’s important to make the difference between a left and a right pseudoinverse.

## Strings in C++ Switch/Case statements

January 10, 2017

Something that used to bug me—used to, because I am so accustomed to work around it that I rarely notice the problem—is that in neither C nor C++ you can use strings (const char * or std::string) in switch/case statement. Indeed, the switch/case statement works only on integral values (an enum, an integral type such as char and int, or an object type with implicit cast to an integral type). But strings aren’t of integral types!

In pure C, we’re pretty much done for. The C preprocessor is too weak to help us built compile-time expression out of strings (or, more exactly, const char *), and there’sn’t much else in the language to help us. However, things are a bit different in C++.

## Logarithms (Part I?)

January 3, 2017

The traditional—but certainly not the best—way to compute the value of the logarithm of some number $x$ is to use a Taylor series, for example

$\displaystyle \ln x = (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\cdots$

but that expansion is only valid for $0, or so, because it is the Taylor expansion of $\ln x$ "around 1", and the convergence radius of this particular expression isn't very large. Furthermore, it needs a great deal of terms before converging.