config/m68k/lb1sf68.S

235 | number, where s=sign, e=exponent, f=fraction. We will call a floating
540 	moveq	IMM (1), d2	/* sign of result stored in d2 (=1 or =-1) */
545 	negb	d2		/* change sign because divisor <0  */
547 	negl	d2		/* change sign because divisor <0  */
727 |  unsigned int sign      : 1;  /* sign bit */
745 	bchg	IMM (31),sp@(12) | change sign of second operand
766 	movel	d0,d7		| get d0's sign bit in d7 '
767 	addl	d1,d1		| check and clear sign bit of a, and gain one
771 	movel	d2,d6		| save sign in d6
772 	addl	d3,d3		| get rid of sign bit and gain one bit of
776 	andl	IMM (0x80000000),d7 | isolate a's sign bit '
777         swap	d6		| and also b's sign bit '
780 	orw	d6,d7		| and combine them into d7, so that a's sign '
1095 	movew	IMM (0),d7	| get a's sign in d7 '
1097 	movew	IMM (0),d6	| and b's sign in d6 '
1119 	andl	IMM (0x80000000),d7 | d7 now has the sign
1202 	exg	d7,a0		| put sign back in a0
1219 	bchg	IMM (31),d7	| change sign bit in d7
1228 	andl	IMM (0x80000000),d7 | isolate sign bit
1282 | Put back the exponent and sign (we don't have overflow). '
1361 	andl	IMM (0x80000000),d7	| Use the sign of a
1398 	orl	d7,d0		| put sign bit back
1432 	movel	d0,d7		| save sign bits
1434 	bclr	IMM (31),d0	| clear sign bits
1452 	eorl	d7,d6		| to check sign bits
1454 	andl	IMM (0x80000000),d7 | get (common) sign bit
1465 	andl	IMM (0x80000000),d7 | get a's sign bit '
1467 	beq	Ld$infty	| if a is INFINITY return with this sign
1469 	bra	Ld$infty	| the opposite sign
1489 	movel	d0,d7			| d7 will hold the sign of the product
1492 	movel	d7,a0			| save sign bit into a0
1496 	bclr	IMM (31),d0		| get rid of a's sign bit '
1501 	bclr	IMM (31),d2		| get rid of b's sign bit '
1512 | Here we have both numbers finite and nonzero (and with no sign bit).
1700 	movel	a0,d7		| get sign bit back into d7
1725 	movel	a0,d7		| get sign bit back into d7
1732 	movel	a0,d7		| get sign bit back into d7
1743 	exg	d3,d1		| and a (with sign bit cleared) into d2-d3
1744 	movel	a0,d0		| set result sign
1753 	movel	a0,d0		| set result sign
1756 	bclr	IMM (31),d2	| clear sign bit
1822 	movel	d0,d7		| d7 will hold the sign of the result
1825 	movel	d7,a0		| save sign into a0
1829 	bclr	IMM (31),d0	| get rid of a's sign bit '
1834 	bclr	IMM (31),d2	| get rid of b's sign bit '
1845 | Here we have both numbers finite and nonzero (and with no sign bit).
1888 | a0	holds the sign of the ratio
2021 	movel	a0,d7		| restore sign bit to d7
2060 | else return +/-INFINITY. Remember that a is in d0 with the sign bit
2062 	movel	a0,d7		| put a's sign bit back in d7 '
2120 | this point the sign of the result is in d7, the result in d0-d1, normalized
2183 	movel	a0,d7		| get back sign bit into d7
2218 | Put back the exponents and sign and return.
2232 	orl	d7,d0		| and sign also
2268 	beq	2f		| if zero (either sign) return +zero
2274 	movel	d0,d7		| else get sign and return INFINITY
2314 | the sign bit.
2638 |  unsigned int sign      : 1;  /* sign bit */
2656 	bchg	IMM (31),sp@(8)	| change sign of second operand
2674 	movel	d0,a0		| get d0's sign bit '
2675 	addl	d0,d0		| check and clear sign bit of a
2677 	movel	d1,a1		| save b's sign bit '
2678 	addl	d1,d1		| get rid of sign bit
2861 	eorl	d6,d7		| combine sign bits
2863 				| sign so we actually subtract the
2874 	movel	a0,d7		| and sign in d7
2945 | We are here if a > 0 and b < 0 (sign bits cleared).
2947 	movel	d6,d7		| put sign in d7
2954 	bchg	IMM (31),d7	| change sign bit in d7
2968 | Now d0-d1 is positive and the sign bit is in d7.
3054 	andl	IMM (0x80000000),d7	| Use the sign of a
3064 	andl	IMM (0x80000000),d7	| put sign in d7
3065 	bclr	IMM (31),d0		| clear sign
3070 	bclr	IMM (31),d7	| if zero be sure to clear sign
3083 	orl	d7,d0		| put sign bit
3099 | Note: when adding two floats of the same sign if either one is
3104 | NaN, but if it is finite we return INFINITY with the corresponding sign.
3113 	movel	d0,d2		| save sign bits
3114 	movel	d0,d7		| into d7 as well as we may need the sign
3117 	bclr	IMM (31),d0	| clear sign bits
3128 	eorl	d3,d2		| to check sign bits
3130 	andl	IMM (0x80000000),d7	| get (common) sign bit
3139 	andl	IMM (0x80000000),d7 | get a's sign bit '
3141 	beq	Lf$infty	| if a is INFINITY return with this sign
3143 	bra	Lf$infty	| the opposite sign
3161 	movel	d0,d7		| d7 will hold the sign of the product
3168 	bclr	IMM (31),d0		| get rid of a's sign bit '
3171 	bclr	IMM (31),d1		| get rid of b's sign bit '
3180 | Here we have both numbers finite and nonzero (and with no sign bit).
3301 | return INFINITY with the correct sign (which is in d7).
3314 1:	bclr	IMM (31),d1	| clear sign bit
3376 	movel	d0,d7			| d7 will hold the sign of the result
3383 	bclr	IMM (31),d0		| get rid of a's sign bit '
3386 	bclr	IMM (31),d1		| get rid of b's sign bit '
3395 | Here we have both numbers finite and nonzero (and with no sign bit).
3435 | d7	holds the sign of the ratio
3479 	movel	a0,d7		| get sign back
3514 	andl	IMM (0x7fffffff),d1	| clear sign bit and test b
3534 | else return +/-INFINITY. Remember that a is in d0 with the sign bit
3655 | Put back the exponents and sign and return.
3669 	orl	d7,d0		| and sign also
3706 	beq	2f		| if zero (either sign) return +zero
3710 	movel	d0,d7		| else get sign and return INFINITY