提交 e5100b32 编写于 作者: S Szabolcs Nagy

math: remove code duplication in erfl found by clang analyzer

erfl had some superflous code left around after the last erf cleanup.
the issue was reported by Alexander Monakov
上级 e71981af
......@@ -266,23 +266,12 @@ static long double erfc2(uint32_t ix, long double x)
s * (ra[5] + s * (ra[6] + s * (ra[7] + s * ra[8])))))));
S = sa[0] + s * (sa[1] + s * (sa[2] + s * (sa[3] + s * (sa[4] +
s * (sa[5] + s * (sa[6] + s * (sa[7] + s * (sa[8] + s))))))));
} else { /* 2.857 <= |x| */
} else if (ix < 0x4001d555) { /* 2.857 <= |x| < 6.6666259765625 */
R = rb[0] + s * (rb[1] + s * (rb[2] + s * (rb[3] + s * (rb[4] +
s * (rb[5] + s * (rb[6] + s * rb[7]))))));
S = sb[0] + s * (sb[1] + s * (sb[2] + s * (sb[3] + s * (sb[4] +
s * (sb[5] + s * (sb[6] + s))))));
}
if (ix < 0x4000b6db) { /* 1.25 <= |x| < 2.85711669921875 ~ 1/.35 */
R = ra[0] + s * (ra[1] + s * (ra[2] + s * (ra[3] + s * (ra[4] +
s * (ra[5] + s * (ra[6] + s * (ra[7] + s * ra[8])))))));
S = sa[0] + s * (sa[1] + s * (sa[2] + s * (sa[3] + s * (sa[4] +
s * (sa[5] + s * (sa[6] + s * (sa[7] + s * (sa[8] + s))))))));
} else if (ix < 0x4001d555) { /* 6.6666259765625 > |x| >= 1/.35 ~ 2.857143 */
R = rb[0] + s * (rb[1] + s * (rb[2] + s * (rb[3] + s * (rb[4] +
s * (rb[5] + s * (rb[6] + s * rb[7]))))));
S = sb[0] + s * (sb[1] + s * (sb[2] + s * (sb[3] + s * (sb[4] +
s * (sb[5] + s * (sb[6] + s))))));
} else { /* 107 > |x| >= 6.666 */
} else { /* 6.666 <= |x| < 107 (erfc only) */
R = rc[0] + s * (rc[1] + s * (rc[2] + s * (rc[3] +
s * (rc[4] + s * rc[5]))));
S = sc[0] + s * (sc[1] + s * (sc[2] + s * (sc[3] +
......
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册