Joseph Ferano
885470820e
(/ 0.0 0.0) printed nan through LLVM, which folds it at compile time to the positive quiet NaN, and -nan through x86, where divsd computes the negative one. Put the operands in globals so nothing folds and both say -nan, so the divergence is the folding path and not the arithmetic. The sign bit of a NaN is not a property of the number and IEEE 754 does not specify it, so the print site is where this is answered. flan_f64_to_bytes renders any NaN as nan, and the two dev emitters do the same. That is not a new rule: format-f64 in the prelude has always answered nan for this value, so a build where (print x) said -nan and (show x 2) said nan was contradicting itself inside one backend. An infinity still prints signed. format.flan prints the three non-finite values through print as well as through show. It is in the survey corpus, so the one program pins the printed form under dune test and the agreement between backends under the survey.
101 lines
4.6 KiB
Plaintext
101 lines
4.6 KiB
Plaintext
;;;; format-f64: a number rendered to a fixed number of decimal places.
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;;;;
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;;;; The runtime's f64->bytes is snprintf "%g" and there is no precision to
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;;;; pass it, so this is the first number formatter in the language that a
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;;;; caller can steer. Every case below is one a plausible wrong version gets
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;;;; wrong, and three of them are the ones that actually ship broken: the
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;;;; carry, where the rounded fraction equals the scale and is not a fraction
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;;;; at all; the zero padding, without which 1.005 prints as "1.5"; and the
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;;;; sign, which belongs to the number and not to its integer part, because
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;;;; -0.5 has an integer part of 0 and 0 carries no sign.
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(defn show [x f64 p i32] ()
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(let [v (format-f64 x p)]
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(println (string (as-slice v)))
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(free v)))
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(defn main [] i32
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;; The ordinary cases, and the one %g cannot do at all: 1/60 wanted to two
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;; places is a frame time, and "%g" answers 0.0166667.
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(show 3.14159 2) ; 3.14
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(show 0.0166667 2) ; 0.02
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(show 1234.5 1) ; 1234.5
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(show 2.0 0) ; 2
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(show 2.0 3) ; 2.000
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;; Rounding is half away from zero at the last digit kept, on both signs.
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(show 0.125 2) ; 0.13
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(show -0.125 2) ; -0.13
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(show 2.5 0) ; 3
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(show -2.5 0) ; -3
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;; The carry. 0.999995 scaled by 10^5 rounds to exactly 100000, which is the
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;; next integer; without the carry this prints "0.100000".
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(show 0.999995 5) ; 1.00000
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(show 9.99 1) ; 10.0
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(show -9.99 1) ; -10.0
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(show 0.99 0) ; 1
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;; Zero padding. The fraction of 1.005 at three places is 5, and five digits
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;; is not the same number as 005.
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(show 1.005 3) ; 1.005
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(show 1.0001 4) ; 1.0001
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(show 7.0 6) ; 7.000000
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;; The sign lives on the number, not on the integer part: both of these have
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;; an integer part of 0, which i64->bytes renders without a sign.
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(show -0.5 2) ; -0.50
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(show -0.004 2) ; -0.00
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;; -0.0 prints as a plain zero. The sign test is (< x 0.0), which -0.0 fails,
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;; and text is not where the sign of a zero should be read from.
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(show 0.0 2) ; 0.00
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(show -0.0 2) ; 0.00
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;; Precision is clamped rather than refused, at both ends.
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(show 1.5 -3) ; 2
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(show 1.5 40) ; 1.500000000
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;; The three inputs with no decimal expansion.
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(show (/ 0.0 0.0) 2) ; nan
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(show (/ 1.0 0.0) 2) ; inf
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(show (/ -1.0 0.0) 2) ; -inf
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;; The same three through print, which goes to the runtime's %g rather than
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;; to format-f64, and the first of them is here for a reason the other two
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;; are not. A NaN carries a sign bit that no arithmetic chose: LLVM folds
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;; (/ 0.0 0.0) at compile time and answers the positive one, divsd answers
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;; the negative one at run time, and "%g" prints the difference. So this
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;; line printed "nan" through one backend and "-nan" through the other for
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;; the same source, and disagreed with the (show ...) above it inside either
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;; one. flan_f64_to_bytes now renders any NaN unsigned, which is what
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;; format-f64 always did. An infinity still prints signed: there the sign is
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;; the value, and both backends were always agreed about it.
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(print (/ 0.0 0.0)) (println "") ; nan
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(print (/ 1.0 0.0)) (println "") ; inf
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(print (/ -1.0 0.0)) (println "") ; -inf
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;; Past 9e18 an f64 has no fractional bits and the integer part does not fit
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;; in an i64, so this falls back to %g rather than approximating.
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(show 1e20 2) ; 1e+20
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;; A large magnitude that does fit, where the fraction is genuinely gone: an
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;; f64 has no bits below 1 up there, so the padding produces the zeros.
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(show 1234567890123.0 2) ; 1234567890123.00
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;; And the thing it is for: a formatted number inside a built string, which
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;; needs the integer part copied out before the fraction is rendered, because
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;; both come through the runtime's one shared scratch buffer.
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(let [b (vec-new u8)]
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(append! (addr b) (bytes "fps "))
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(let [f (format-f64 59.94 1)]
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(append! (addr b) (as-slice f))
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(free f))
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(append! (addr b) (bytes " / frame "))
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(let [f (format-f64 0.0166667 4)]
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(append! (addr b) (as-slice f))
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(free f))
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(println (string (as-slice b))) ; fps 59.9 / frame 0.0167
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(free b))
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0)
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