Format double in decimal notation without trailing zeros after the decimal point

[akaDemik via Digitalmars-d-learn]

Thanks for the reply.
I remember about the accuracy of floating point numbers.
It is encouraging that the "%g" can handle it.
format("%.17g", 123456.789123); // == 123456.789123
And we have a flag "#". As mentioned in documentation:
> '#' floating Always insert the decimal point and print trailing 
> zeros.
With '#' I get:
format("%#.17g", 123456.789123); // == 123456.78912300000
So, with the flag '#' "%g" is almost similar to the "%f".
But there is no flag repealing '#' if it is enabled by default 
(in the case of "%f").
I looked deeper into the function format and realized that my 
question is more related to the implementation of snprintf. If I 
understand correctly, snprintf does the job, and std.format 
provides a safe and very convenient wrapper.

On Friday, 27 March 2015 at 17:08:07 UTC, Steven Schveighoffer 
wrote:
> On 3/27/15 11:02 AM, akaDemik wrote:
>> The task seemed very simple. But I'm stuck.
>> I want to:
>>   1234567890123.0 to "1234567890123"
>>   1.23 to "1.23"
>>   1.234567 to "1.2346".
>> With format string "%.4f" i get "1.2300" for 1.23.
>> With "%g" i get "1.23456789e+12" for "1234567890123.0".
>> I can not believe that it is not implemented. What did I miss?
>
> I think you are asking for trouble to do this. Floating point 
> is not exact, so for example, if I do
>
> writefln("%.15f", 123456.789123);
>
> I get:
>
> 123456.789122999995016
>
> How far do you want to go before you determine there will only 
> be zeros? It could be infinity.
>
> I'd say your best bet is to format to the max level you want, 
> e.g. "%.6f"
>
> Then trim off any trailing zeros (and decimal point if 
> necessary) after conversion to a string.
>
> -Steve