为FMOD（）确切当y是一个整数？ [英] Is fmod() exact when y is an integer?
问题描述
在使用双FMOD（双X，双Y）和是是一个整数，结果出现是的总是的准确。
In using double fmod(double x, double y) and y is an integer, the result appears to be always exact.
（即是整体确切的数字，没有意思 INT 在这里。）
(That is y a whole exact number, not meaning int here.)
也许C没有的需要的 FMOD（）来提供这些选择的情况下一个确切的答案，但对编译器我已经试过了，结果是准确的，甚至当 X / Y 是不完全的重新presentable。
Maybe C does not require fmod() to provide an exact answers in these select cases, but on compilers I've tried, the result is exact, even when the quotient of x/y is not exactly representable.
- 预计确切的答案时，
是是一个整数？ - 如果不是，请提供一个反例。
- Are exact answers expected when
yis an integer? - If not, please supply a counter example.
例如：
double x = 1e10;
// x = 10000000000
printf("%.50g\n", fmod(x, 100));
// prints 0
x = 1e60;
// x = 999999999999999949387135297074018866963645011013410073083904
printf("%.50g\n", fmod(x, 100));
// prints 4
x = DBL_MAX;
// x = 179769313486231570...6184124858368
printf("%.50g\n", fmod(x, 100));
// prints 68
x = 123400000000.0 / 9999;
// x = 12341234.1234123408794403076171875
printf("%.50g %a\n", fmod(x, 100), fmod(x, 100));
// prints 34.1234123408794403076171875 0x1.10fcbf9cp+5
注：结果
我的双击出现在IEEE 754 binary64标准。结果的printf的限制（）不成问题在这里，只需 FMOD（）。
Notes:
My double appears to the IEEE 754 binary64 compliant.
The limitations of printf() are not at issue here, just fmod().
注：按是准确的答案预期，我问，如果在 FMOD（）结果和数学结果是完全一样的。
Note: By "Are exact answers expected", I was asking if the the fmod() result and the mathematical result are exactly the same.
推荐答案
在IEEE 754标准定义，其余操作 X / Y ，圆（X / Y）等。有不精确重新presentations。
The IEEE Standard 754 defines the remainder operation x REM y as the mathematical operation x - (round(x/y)*y). The result is exact by definition, even when the intermediate operations x/y, round(x/y), etc. have inexact representations.
正如aka.nice指出，上面的定义匹配的libm 库函数剩余。 FMOD 以不同的方式被定义，要求其结果具有相同的符号 X 。然而，由于之间的差异 FMOD 和剩余是 0 或是，我认为，这仍然解释了为什么结果是准确的。
As pointed out by aka.nice, the definition above matches the library function remainder in libm. fmod is defined in a different way, requiring that the result has the same sign as x. However, since the difference between fmod and remainder is either 0 or y, I believe that this still explains why the result is exact.
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