diff options
Diffstat (limited to 'gcc/config/m68k/lb1sf68.asm')
| -rw-r--r-- | gcc/config/m68k/lb1sf68.asm | 44 |
1 files changed, 22 insertions, 22 deletions
diff --git a/gcc/config/m68k/lb1sf68.asm b/gcc/config/m68k/lb1sf68.asm index 922df8f8a3e..7833e5388d2 100644 --- a/gcc/config/m68k/lb1sf68.asm +++ b/gcc/config/m68k/lb1sf68.asm @@ -315,7 +315,7 @@ L4: lsrl IMM (1), d1 /* shift divisor */ divu d1, d0 /* now we have 16 bit divisor */ andl IMM (0xffff), d0 /* mask out divisor, ignore remainder */ -/* Muliply the 16 bit tentative quotient with the 32 bit divisor. Because of +/* Multiply the 16 bit tentative quotient with the 32 bit divisor. Because of the operand ranges, this might give a 33 bit product. If this product is greater than the dividend, the tentative quotient was too large. */ movel d2, d1 @@ -628,7 +628,7 @@ Ladddf$2: | Here we have a's exponent larger than b's, so we have to shift b. We do | this by using as counter d2: 1: movew d4,d2 | move largest exponent to d2 - subw d5,d2 | and substract second exponent + subw d5,d2 | and subtract second exponent exg d4,a2 | get back the longs we saved exg d5,a3 | | if difference is too large we don't shift (actually, we can just exit) ' @@ -714,7 +714,7 @@ Ladddf$4: | Now we have the numbers in d0--d3 and d4--d7, the exponent in a2, and | the signs in a4. -| Here we have to decide whether to add or substract the numbers: +| Here we have to decide whether to add or subtract the numbers: exg d7,a0 | get the signs exg d6,a3 | a3 is free to be used movel d7,d6 | @@ -723,7 +723,7 @@ Ladddf$4: movew IMM (0),d6 | and b's sign in d6 ' eorl d7,d6 | compare the signs bmi Lsubdf$0 | if the signs are different we have - | to substract + | to subtract exg d7,a0 | else we add the numbers exg d6,a3 | addl d7,d3 | @@ -739,7 +739,7 @@ Ladddf$4: | Before rounding normalize so bit #DBL_MANT_DIG is set (we will consider | the case of denormalized numbers in the rounding routine itself). -| As in the addition (not in the substraction!) we could have set +| As in the addition (not in the subtraction!) we could have set | one more bit we check this: btst IMM (DBL_MANT_DIG+1),d0 beq 1f @@ -772,7 +772,7 @@ Ladddf$5: bra Ld$overflow Lsubdf$0: -| Here we do the substraction. +| Here we do the subtraction. exg d7,a0 | put sign back in a0 exg d6,a3 | subl d7,d3 | @@ -796,7 +796,7 @@ Lsubdf$0: | Before rounding normalize so bit #DBL_MANT_DIG is set (we will consider | the case of denormalized numbers in the rounding routine itself). -| As in the addition (not in the substraction!) we could have set +| As in the addition (not in the subtraction!) we could have set | one more bit we check this: btst IMM (DBL_MANT_DIG+1),d0 beq 1f @@ -927,7 +927,7 @@ Ladddf$nf: 3: | Now comes the check for +/-INFINITY. We know that both are (maybe not | finite) numbers, but we have to check if both are infinite whether we -| are adding or substracting them. +| are adding or subtracting them. eorl d7,d6 | to check sign bits bmi 1f andl IMM (0x80000000),d7 | get (common) sign bit @@ -999,7 +999,7 @@ Lmuldf$1: lsrw IMM (4),d5 | Lmuldf$2: | addw d5,d4 | add exponents - subw IMM (D_BIAS+1),d4 | and substract bias (plus one) + subw IMM (D_BIAS+1),d4 | and subtract bias (plus one) | We are now ready to do the multiplication. The situation is as follows: | both a and b have bit 52 ( bit 20 of d0 and d2) set (even if they were @@ -1210,7 +1210,7 @@ Ldivdf$1: | swap d5 | lsrw IMM (4),d5 | Ldivdf$2: | - subw d5,d4 | substract exponents + subw d5,d4 | subtract exponents addw IMM (D_BIAS),d4 | and add bias | We are now ready to do the division. We have prepared things in such a way @@ -1245,7 +1245,7 @@ Ldivdf$2: | dbra d5,1b | and branch back bra 5f 4: cmpl d1,d3 | here d0==d2, so check d1 and d3 - bhi 3b | if d1 > d2 skip the substraction + bhi 3b | if d1 > d2 skip the subtraction bra 2b | else go do it 5: | Here we have to start setting the bits in the second long. @@ -1262,7 +1262,7 @@ Ldivdf$2: | dbra d5,1b | and branch back bra 5f 4: cmpl d1,d3 | here d0==d2, so check d1 and d3 - bhi 3b | if d1 > d2 skip the substraction + bhi 3b | if d1 > d2 skip the subtraction bra 2b | else go do it 5: | Now go ahead checking until we hit a one, which we store in d2. @@ -1902,12 +1902,12 @@ Laddsf$2: | signs are stored in a0 and a1). Laddsf$3: -| Here we have to decide whether to add or substract the numbers +| Here we have to decide whether to add or subtract the numbers exg d6,a0 | get signs back exg d7,a1 | and save the exponents eorl d6,d7 | combine sign bits bmi Lsubsf$0 | if negative a and b have opposite - | sign so we actually substract the + | sign so we actually subtract the | numbers | Here we have both positive or both negative @@ -1926,7 +1926,7 @@ Laddsf$3: | Before rounding normalize so bit #FLT_MANT_DIG is set (we will consider | the case of denormalized numbers in the rounding routine itself). -| As in the addition (not in the substraction!) we could have set +| As in the addition (not in the subtraction!) we could have set | one more bit we check this: btst IMM (FLT_MANT_DIG+1),d0 beq 1f @@ -1957,7 +1957,7 @@ Laddsf$4: Lsubsf$0: | We are here if a > 0 and b < 0 (sign bits cleared). -| Here we do the substraction. +| Here we do the subtraction. movel d6,d7 | put sign in d7 andl IMM (0x80000000),d7 @@ -1974,7 +1974,7 @@ Lsubsf$0: | Now d0-d1 is positive and the sign bit is in d7. -| Note that we do not have to normalize, since in the substraction bit +| Note that we do not have to normalize, since in the subtraction bit | #FLT_MANT_DIG+1 is never set, and denormalized numbers are handled by | the rounding routines themselves. lea Lsubsf$1,a0 | to return from rounding routine @@ -2071,7 +2071,7 @@ Laddsf$ret$den: | Note: when adding two floats of the same sign if either one is | NaN we return NaN without regard to whether the other is finite or -| not. When substracting them (i.e., when adding two numbers of +| not. When subtracting them (i.e., when adding two numbers of | opposite signs) things are more complicated: if both are INFINITY | we return NaN, if only one is INFINITY and the other is NaN we return | NaN, but if it is finite we return INFINITY with the corresponding sign. @@ -2095,7 +2095,7 @@ Laddsf$nf: bhi Lf$inop | Now comes the check for +/-INFINITY. We know that both are (maybe not | finite) numbers, but we have to check if both are infinite whether we -| are adding or substracting them. +| are adding or subtracting them. eorl d3,d2 | to check sign bits bmi 1f movel d0,d7 @@ -2161,7 +2161,7 @@ Lmulsf$1: | number lsrw IMM (7),d3 | Lmulsf$2: | addw d3,d2 | add exponents - subw IMM (F_BIAS+1),d2 | and substract bias (plus one) + subw IMM (F_BIAS+1),d2 | and subtract bias (plus one) | We are now ready to do the multiplication. The situation is as follows: | both a and b have bit FLT_MANT_DIG-1 set (even if they were @@ -2292,7 +2292,7 @@ SYM (__divsf3): beq Ldivsf$b$0 | branch if b is zero cmpl d6,d0 | is a big? bhi Ldivsf$inop | if a is NaN return NaN - beq Ldivsf$inf | if a is INIFINITY we have to check b + beq Ldivsf$inf | if a is INFINITY we have to check b cmpl d6,d1 | now compare b with INFINITY bhi Ldivsf$inop | if b is NaN return NaN beq Ldivsf$underflow @@ -2314,7 +2314,7 @@ Ldivsf$1: | swap d3 | lsrw IMM (7),d3 | Ldivsf$2: | - subw d3,d2 | substract exponents + subw d3,d2 | subtract exponents addw IMM (F_BIAS),d2 | and add bias | We are now ready to do the division. We have prepared things in such a way |