Re: [bug-gawk] How to print the just enough number digits so the machine representation of the number does not change?

[Eli Zaretskii]

> Date: Mon, 8 Oct 2012 06:03:10 -0600 (MDT)
> From: "Nelson H. F. Beebe" <address@hidden>
> Cc: address@hidden, Peng Yu <address@hidden>, address@hidden
> 
> I beg to differ on the number of digits required for binary to decimal
> conversion.

Sorry, but I don't understand the bottom line.  What does this mean in
terms of answer to the OP's question?  Is my suggestion to use %.20g
incorrect, and if so, what is the correct answer?

> hoc> matula(53) 
>  17

That's what I said, IMO.

> hoc> goldberg(16)
>  55
> 
> Notice that the two functions are not mutually symmetric: 16 decimal digits
> needs two more bits than the 64-bit binary format can supply.

Isn't one of these bits hidden?

More importantly, I still don't see the practical consequences.
Suppose I used 20-digit format to print a 'double', then read it with
%f -- are you saying the extra bits will somehow affect the machine
representation of the result after reading?  If so, how, and could you
show an example of this, please?

> The default output format in decimal for a 32-bit IEEE binary type should be
> 9 digits.  The 32-bit IEEE decimal type needs 25 bits for correct round-trip
> conversion.

The OP asked (albeit implicitly) about 'double', not 'float'.

Thanks.