Why is not the output precision being rounded correctly?

Question

I have tested the following code to verify some simple numerical outputs when std::setprecision is used in C++:

double f1 = 3.1415;
std::cout << std::setprecision(4) << f1 << '\n';
double f2 = 1.2345;
std::cout << std::setprecision(4) << f2 << '\n';

The corresponding outputs are:

3.142
1.234

Why is 1.234 not being rounded to 1.235? Furthermore, if we change f2 to 1.2315, then the rounding in the output does happen, and the output is 1.232. Is there any rule of thumb to follow when using setprecision to format the output of numeric values?

`f2` is probably actually represented by `1.234499999999` or something like that, so it gets rounded down. — François Andrieux, Feb 12 '20 at 21:30
@FrançoisAndrieux It does makes sense if `f2` is represented as you noted! — 영민 카이 앤절, Feb 12 '20 at 21:33
https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html — Jesper Juhl, Feb 12 '20 at 21:33
I think an odd digit followed by 5 gets rounded up, while an even digit gets rounded down — Krzysiek Karbowiak, Feb 12 '20 at 21:33
https://stackoverflow.com/questions/20264681/how-to-specify-setprecision-rounding — Vlad Feinstein, Feb 12 '20 at 21:38

score 1 · Accepted Answer · answered Feb 12 '20 at 22:06

The comment tells you why. I strongly recommend to read https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

In order to do correct rounding, C++ comes with round() for that but it only round to integer, not arbitrary decimal places. The following can give you 1.235, although a bit ugly:

double f1 = 3.1415;
std::cout << std::setprecision(4) << round(f1*1000)/1000 << '\n';
double f2 = 1.2345;
std::cout << std::setprecision(4) << round(f2*1000)/1000 << '\n';

Why is not the output precision being rounded correctly?

1 Answers1