Merge from HEAD

; view, i.e. when the highest bit is set), dividing the dividend by

; 2 isn't enough, it needs to be divided by 4.  Furthermore, the

; divisor needs to be divided by 2 (unsigned) as well, to avoid more

; problems with the sign.  In this case, the divisor is so large,

; from an unsigned point of view, that the dropped lowest bit is

; insignificant for the operation, and therefore doesn't need

; bothering with.  The remainder might end up incorrect, bit that's

; adjusted at the end of the routine anyway.

; problems with the sign.  In this case, a little extra fiddling with

; the remainder is required.

;

; So, the simplest way to handle this is always to divide the dividend

; by 4, and to divide the divisor by 2 if it's highest bit is set.

; if (q < 0) q = -q		# I doubt this is necessary any more

;

; r'    = r >> 30

; if (d' > 0) q = q << 1

; if (d' >= 0) q = q << 1

; q     = q << 1

; r     = (r << 2) + l'

;

; if (d' < 0)

;   {

;     [r',r] = [r',r] - q

;     while ([r',r] < 0)

;       {

;         [r',r] = [r',r] + d

;         q = q - 1

;       }

;   }

;

; while ([r',r] >= d)

;   {

;     [r',r] = [r',r] - d

	bicl3	#^X00000003,r3,r3

	addl	r5,r3		; r = r + l'

	tstl	r7

	bgeq	5$

	bitl	#1,r7

	beql	5$		; if d < 0 && d & 1

	subl	r2,r3		;   [r',r] = [r',r] - q

	sbwc	#0,r6

45$:

	bgeq	5$		;   while r < 0

	decl	r2		;     q = q - 1

	addl	r7,r3		;     [r',r] = [r',r] + d

	adwc	#0,r6

	brb	45$

5$:

	tstl	r6

	bneq	6$

	cmpl	r3,r7

	blssu	42$		; while [r',r] >= d'

6$:

	subl	r7,r3		;   r = r - d

	subl	r7,r3		;   [r',r] = [r',r] - d

	sbwc	#0,r6

	incl	r2		;   q = q + 1

	brb	5$