Date: Sun, 5 Sep 2004 21:51:52 +0200 From: Erik Trulsson <ertr1013@student.uu.se> To: Steve Kargl <sgk@troutmask.apl.washington.edu> Cc: freebsd-standards@freebsd.org Subject: Re: is printf() broken? Message-ID: <20040905195152.GA6096@falcon.midgard.homeip.net> In-Reply-To: <20040905193640.GA37190@troutmask.apl.washington.edu> References: <20040905193640.GA37190@troutmask.apl.washington.edu>
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On Sun, Sep 05, 2004 at 12:36:40PM -0700, Steve Kargl wrote:
> The following program
>
> #include <stdio.h>
> int main(void) {
> int d;
> double x;
> x = 1.234E05;
> for (d = 0; d < 5; d++)
> printf("%+-31.*e\n", d, x);
> return 0;
> }
>
> generates
>
> +1e+05
> +1.2e+05
> +1.23e+05
> +1.234e+05
> +1.2340e+05
>
> The question is whether the first number should be
> "+1.e+05". That is, is the printing of the decimal
> point required or optional? I only have Harbison
> and Steele's book and it does not state what the
> expected behavior should be.
I believe printf() is correct, and that the printing of the decimal
point for the first number above is neither required nor optional but
actually not allowed at all
>From Harbison & Steele 5th edition:
The e and E conversions
[...]
If the precision is 0, then no digits appear after the decimal point.
Moreover the decimal point also does not appear unless the # flag is
present.
(Nearly identical text is also in H&S 3rd edition.)
>From n869.txt (the last publically available draft of C99)
[the e and E conversions for fprintf]
[...]
if the precision is zero and the # flag is not
specified, no decimal-point character appears
--
<Insert your favourite quote here.>
Erik Trulsson
ertr1013@student.uu.se
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