|
inlineconstexpr |
Parameters
x
, y
: floating values.c
: Conditional expression masking the operation.m
: Logical value masking the operation.Return value
the arc tangent of \(\frac{y}x\) in \(\pi\), in the range [-1, +1], is returned. The IEEE limiting values are almost all satisfied :
x
and y
are both zero or infinites, Nan is returned (this is not standard conforming)y
is \(\pm0\) and x
is strictly negative or \(-0\), \(\pm1\) is returnedy
is \(\pm0\) and x
is strictly positive or \(+0\), \(\pm0\) is returnedy
is \(\pm\infty\) and x
is finite, \(\pm\frac12\) is returnedx
is \(\pm0\) and y
is strictly negative, \(-\frac12\) is returnedx
is \(\pm0\) and y
is strictly positive, \(+\frac12\) is returnedx
is \(-\infty\) and y
is finite and positive, \(+1\) is returnedx
is \(-\infty\) and y
is finite and negative, \(-1\) is returnedx
is \(+\infty\) and y
is finite and positive, \(+0\) is returnedx
is \(+\infty\) and y
is finite and negative, \(-0\) is returnedx
is Nan or y
is Nan, Nan is returnedThe call will return a NaN if x
and y
are both either null or infinite: this result is not IEEE conformant, but allows to simplify (and speed) the implementation. In all other cases, the result is standard conformant.
y
is \(\pm\infty\) and x
is \(-\infty\), \(\pm\frac34\) is returnedy
is \(\pm\infty\) and x
is \(+\infty\), \(\pm\frac14\) is returnedx
is \(\pm0\) and y
is \(\pm-0\), \(-\frac12\) is returnedx
is \(\pm0\) and y
is \(\pm+0\), \(+\frac12\) is returned