+ 6

# <limits>, long long doubles and precision

Did you ever work with very little / very large numbers? How can you output a 123! or to be sure your approximation of pi to 57-th digit is not mispelled by the compiler? Feel free to post your (also not-C++) codes.

10th Feb 2019, 10:19 PM
Paolo De Nictolis
+ 2
One solution is to split the long number over several smaller numbers, like for example here: https://code.sololearn.com/cV3o3NWxVZ11/?ref=app And about correct decimals: Do it like a kid at school, (int) digit by digit: https://code.sololearn.com/cnCk4wRLTkS6/?ref=app Problem with both approaches: Although the output may be right, they are not number types like you know them anymore; you'd have to write a complete math around them.
10th Feb 2019, 10:23 PM
HonFu
+ 3
HonFu all right, but in your code you're not generous with comments and samples!
10th Feb 2019, 10:42 PM
Paolo De Nictolis
+ 3
That's right... I'm just now starting to get into the habit of writing comments... for now I can offer you to ask questions. :-) But samples? There are samples! Run the factorial code and see all these tapeworm numbers! And rooty gives you square root of 2; for other numbers you just have to change the argument.
10th Feb 2019, 10:49 PM
HonFu
+ 3
I can try to comment a bit after the fact. You remember how we multiply larger numbers in school, right? 123*5 First we take the 3*5. If it's larger than ten, we just take the last digit and keep the other digit by adding it to the next digits' result. So 15 -> 5 and we keep 1 and add it to the 10 (2*5), keep the 1... After we are through, we got the complete result. In programming languages, you get the last digit by %10; then you can calculate 15/10 (// in Python) and get 1 (what you keep). Now imagine we don't do this with %10, but with %1'000'000'000. You'd get bigger chunks, but actually it works quite the same (I needed a little help from a friend who can actually do maths for this). Largest number you could get would be 999'999'999*999'999'999 which barely fits into an unsigned long long int. Now if you do this 'digit' by 'digit', you can store each of them in a list (vector in this case). Then you basically just have to put them out and it looks like one large number.
10th Feb 2019, 11:14 PM
HonFu