Meta: with regards to significant digits, it may depend on application, but this article reminded me on NASA's 'take' on π (pi):
> To start, let me answer your question directly. For JPL's highest accuracy calculations, which are for interplanetary navigation, we use 3.141592653589793. Let's look at this a little more closely to understand why we don't use more decimal places. […]
> 3. Let's go to the largest size there is: the known universe. The radius of the universe is about 46 billion light years. Now let me ask (and answer!) a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom, the simplest atom? It turns out that 37 decimal places (38 digits, including the number 3 to the left of the decimal point) would be quite sufficient. […]
Meta: with regards to significant digits, it may depend on application, but this article reminded me on NASA's 'take' on π (pi):
> To start, let me answer your question directly. For JPL's highest accuracy calculations, which are for interplanetary navigation, we use 3.141592653589793. Let's look at this a little more closely to understand why we don't use more decimal places. […]
> 3. Let's go to the largest size there is: the known universe. The radius of the universe is about 46 billion light years. Now let me ask (and answer!) a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom, the simplest atom? It turns out that 37 decimal places (38 digits, including the number 3 to the left of the decimal point) would be quite sufficient. […]
* https://www.jpl.nasa.gov/edu/news/how-many-decimals-of-pi-do...
For a short moment I got excited he may have started again… this post needs a [2019] :-(
Fixed
tcc also supports a binary float extension. https://github.com/TinyCC/tinycc/blob/9b8765d8baaeb2a16112d6...
long double la0 = 0B.110101100P12L;