mirror of https://github.com/postgres/postgres
Windows has this since _MSC_VER >= 1200, and so do all other live platforms according to the buildfarm, so remove the configure probe and src/port/ substitution. This is part of a series of commits to get rid of no-longer-relevant configure checks and dead src/port/ code. I'm committing them separately to make it easier to back out individual changes if they prove less portable than I expect. Discussion: https://postgr.es/m/15379.1582221614@sss.pgh.pa.uspull/51/head
Tom Lane
6 years ago
parent
1200d71a09
commit
f88a058200
@ -1,97 +0,0 @@ |
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/*-------------------------------------------------------------------------
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* |
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* rint.c |
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* rint() implementation |
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* |
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* By Pedro Gimeno Fortea, donated to the public domain |
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* |
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* IDENTIFICATION |
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* src/port/rint.c |
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* |
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*------------------------------------------------------------------------- |
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*/ |
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#include "c.h" |
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#include <math.h> |
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/*
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* Round to nearest integer, with halfway cases going to the nearest even. |
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*/ |
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double |
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rint(double x) |
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{ |
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double x_orig; |
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double r; |
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/* Per POSIX, NaNs must be returned unchanged. */ |
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if (isnan(x)) |
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return x; |
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if (x <= 0.0) |
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{ |
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/* Both positive and negative zero should be returned unchanged. */ |
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if (x == 0.0) |
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return x; |
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/*
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* Subtracting 0.5 from a number very close to -0.5 can round to |
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* exactly -1.0, producing incorrect results, so we take the opposite |
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* approach: add 0.5 to the negative number, so that it goes closer to |
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* zero (or at most to +0.5, which is dealt with next), avoiding the |
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* precision issue. |
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*/ |
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x_orig = x; |
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x += 0.5; |
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/*
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* Be careful to return minus zero when input+0.5 >= 0, as that's what |
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* rint() should return with negative input. |
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*/ |
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if (x >= 0.0) |
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return -0.0; |
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/*
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* For very big numbers the input may have no decimals. That case is |
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* detected by testing x+0.5 == x+1.0; if that happens, the input is |
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* returned unchanged. This also covers the case of minus infinity. |
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*/ |
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if (x == x_orig + 1.0) |
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return x_orig; |
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/* Otherwise produce a rounded estimate. */ |
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r = floor(x); |
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/*
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* If the rounding did not produce exactly input+0.5 then we're done. |
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*/ |
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if (r != x) |
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return r; |
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/*
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* The original fractional part was exactly 0.5 (since |
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* floor(input+0.5) == input+0.5). We need to round to nearest even. |
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* Dividing input+0.5 by 2, taking the floor and multiplying by 2 |
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* yields the closest even number. This part assumes that division by |
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* 2 is exact, which should be OK because underflow is impossible |
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* here: x is an integer. |
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*/ |
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return floor(x * 0.5) * 2.0; |
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} |
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else |
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{ |
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/*
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* The positive case is similar but with signs inverted and using |
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* ceil() instead of floor(). |
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*/ |
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x_orig = x; |
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x -= 0.5; |
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if (x <= 0.0) |
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return 0.0; |
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if (x == x_orig - 1.0) |
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return x_orig; |
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r = ceil(x); |
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if (r != x) |
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return r; |
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return ceil(x * 0.5) * 2.0; |
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} |
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} |
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