Date: Sun, 5 Sep 2004 21:56:38 +0200 From: Stefan Farfeleder <stefan@fafoe.narf.at> To: Steve Kargl <sgk@troutmask.apl.washington.edu> Cc: freebsd-standards@freebsd.org Subject: Re: is printf() broken? Message-ID: <20040905195632.GG72545@wombat.fafoe.narf.at> 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.
The output is correct:
e, E A double argument representing a floating-point number is converted in the
style [-]d.ddd eħdd, where there is one digit (which is nonzero if the
argument is nonzero) before the decimal-point character and the number of
digits after it is equal to the precision; if the precision is missing, it is taken as
6; if the precision is zero and the # flag is not specified, no decimal-point
character appears. The value is rounded to the appropriate number of digits.
[...]
Cheers,
Stefan
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