15  *   1. Argument Reduction: find k and f such that
16  *			1+x = 2^k * (1+f),
19  *      Note. If k=0, then f=x is exact. However, if k!=0, then f
28  *		 = 2s + 2/3 s**3 + 2/5 s**5 + .....,
46  *	3. Finally, log1p(x) = k*ln2 + log1p(f).
47  *		 	     = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
83 ln2_hi  =  6.93147180369123816490e-01,	/* 3fe62e42 fee00000 */
84 ln2_lo  =  1.90821492927058770002e-10,	/* 3dea39ef 35793c76 */
86 Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
87 Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
88 Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
89 Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
90 Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
91 Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
92 Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
101 	int32_t k,hx,hu,ax;  in log1p()  local
106 	k = 1;  in log1p()
120 		k=0;f=x;hu=1;}		/* sqrt(2)/2- <= 1+x < sqrt(2)+ */  in log1p()
123 	if(k!=0) {  in log1p()
127 	        k  = (hu>>20)-1023;  in log1p()
128 	        c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */  in log1p()
133 	        k  = (hu>>20)-1023;  in log1p()
140 	     * strict bounds than the one here so that the k==0 case is  in log1p()
142 	     * using the correction term but don't use it if k==0.  in log1p()
147 	        k += 1;  in log1p()
156 		if(k==0) {  in log1p()
159 		    c += k*ln2_lo;  in log1p()
160 		    return k*ln2_hi+c;  in log1p()
164 	    if(k==0) return f-R; else  in log1p()
165 	    	     return k*ln2_hi-((R-(k*ln2_lo+c))-f);  in log1p()
170 	if(k==0) return f-(hfsq-s*(hfsq+R)); else  in log1p()
171 		 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);  in log1p()