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58  * Our task is to calculate the square root of a floating point number x0.
127  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
133  * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
135  * value, for which q is some power of two times its square root, x0.)
140  *	q = y = 0; x = x0;
156  * If x0 is fixed point, rather than an integer, we can simply alter the
157  * scale factor between q and sqrt(x0).  As it happens, we can easily arrange
158  * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
160  * In our case, however, x0 (and therefore x, y, q, and t) are multiword
170  * intermediate calculations can overflow.  We know that x0 is in [1..4)
193 	u_int x0, x1, x2, x3;
225 	FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
229 	x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
240 	x0 = x->fp_mant[0];
274 		x0 -= bit;
281 		if (x0 >= t0) {		/* if x >= t then */
282 			x0 -= t0;	/*	x -= t */
300 	FPU_SUBC(d0, x0, t0);		/* d = x - t */
302 		x0 = d0, x1 = d1;	/*	x -= t */
311 		FPU_SUBC(d0, x0, t0);
313 			x0 = d0, x1 = d1;
332 	FPU_SUBC(d0, x0, t0);
334 		x0 = d0, x1 = d1, x2 = d2;
344 		FPU_SUBC(d0, x0, t0);
346 			x0 = d0, x1 = d1, x2 = d2;
366 	FPU_SUBC(d0, x0, t0);
369 		x0 = d0, x1 = d1, x2 = d2;
379 		FPU_SUBC(d0, x0, t0);
381 			x0 = d0, x1 = d1, x2 = d2;
393 	x->fp_sticky = x0 | x1 | x2 | x3;