Removing leading and trailing zeros from %f in C++

Question

What is the best way to remove leading and trailing zeros from C++ string formatters that use %f. There's a similar question Avoid trailing zeroes in printf() but that just says to use the %g formatter.

The %g formatter will output 600000000009 as 6e+11

I just want to remove the leading and trailing zeros from %f eg:

000.001 -> 0.001
00.100 -> 0.100
.10 -> 0.1
2.0 -> 2

Any thoughts/suggestions welcome. Thanks.

shouldn't `.10` go to `0.1` ??? – Tomer W Jul 20 '15 at 17:21 — Tomer W, Jul 20 '15 at 17:21
Yeah you're right. sorry just made the edit – Adroit Jul 20 '15 at 17:24 — Adroit, Jul 20 '15 at 17:24

6502 · Accepted Answer · 2015-07-21T16:39:53.833

2

A dynamic format that works in most cases is just

%.15g

that means up to 15 digits of precision.

An even better approach is to start from 15 digits and check back if atof(result) is the same as the input, incrementing the number of digits until it does match.

Using directly the maximum number of digits can be a bad idea because formatting for example 0.1 with 18 digits will bring out a representation that will look strange for the user.

edited Jul 21 '15 at 16:39

answered Jul 20 '15 at 17:23

6502

112,025
15
165
265

This disagrees with the accepted answer on the cited question. Who's right? Can you provide some _evidence_ for your claim? – Lightness Races in Orbit Jul 20 '15 at 17:46
Not sure what you mean. The answer is correct it works in most cases unless you have different standards of precision in you apps. – Adroit Jul 21 '15 at 16:03

Removing leading and trailing zeros from %f in C++

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