* Copyright (c) 1980 Regents of the University of California.
* Redistribution and use in source and binary forms are permitted
* provided that the above copyright notice and this paragraph are
* duplicated in all such forms and that any documentation,
* advertising materials, and other materials related to such
* distribution and use acknowledge that the software was developed
* by the University of California, Berkeley. The name of the
* University may not be used to endorse or promote products derived
* from this software without specific prior written permission.
* THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR
* IMPLIED WARRANTIES, INCLUDING, WITHOUT LIMITATION, THE IMPLIED
* WARRANTIES OF MERCHANTIBILITY AND FITNESS FOR A PARTICULAR PURPOSE.
#if defined(LIBC_SCCS) && !defined(lint)
.asciz "@(#)atof.s 5.5 (Berkeley) 6/27/88"
#endif /* LIBC_SCCS and not lint */
* atof: convert ascii to floating
* r0-1: value being developed
* r2: first section: pointer to the next character
* second section: binary exponent
* r4: first section: the current character
* second section: scratch
* r5: the decimal exponent
.set msign,0 # mantissa has negative sign
.set esign,1 # exponent has negative sign
.set decpt,2 # decimal point encountered
clrl r3 # All flags start out false
movl 4(ap),r2 # Address the first character
clrl r5 # Clear starting exponent
* Skip leading white space
sk0: movzbl (r2)+,r4 # Fetch the next (first) character
cmpb $' ,r4 # Is it blank?
cmpb r4,$8 # 8 is lowest of white-space group
jlss sk1 # Jump if char too low to be white space
cmpb r4,$13 # 13 is highest of white-space group
jleq sk0 # Jump if character is white space
cmpb $'+,r4 # Positive sign?
cmpb $'-,r4 # Negative sign?
bisb2 $1<msign,r3 # Indicate a negative mantissa
cs1: movzbl (r2)+,r4 # Skip the character
* Accumulate digits, keeping track of the exponent
clrq r0 # Clear the accumulator
ad0: cmpb r4,$'0 # Do we have a digit?
jlss ad4 # ... no, too small
jgtr ad4 # ... no, too large
* We got a digit. Accumulate it
cmpl r1,$214748364 # Would this digit cause overflow?
* Multiply (r0,r1) by 10. This is done by developing
* (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits,
* and adding the two quadwords.
ashq $1,r0,r6 # (r6,r7)=(r0,r1)*2
ashq $3,r0,r0 # (r0,r1)=(r0,r1)*8
addl2 r6,r0 # Add low halves
adwc r7,r1 # Add high halves
subl2 $'0,r4 # Get the digit value
addl2 r4,r0 # Add it into the accumulator
adwc $0,r1 # Possible carry into high half
jbr ad2 # Join common code
* Here when the digit won't fit in the accumulator
ad1: incl r5 # Ignore the digit, bump exponent
* If we have seen a decimal point, decrease the exponent by 1
ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen
decl r5 # Decrease exponent
* Fetch the next character, back for more
* Not a digit. Could it be a decimal point?
ad4: cmpb r4,$'. # If it's not a decimal point, either it's
jneq ad5 # the end of the number or the start of
jbcs $decpt,r3,ad3 # If it IS a decimal point, we record that
# we've seen one, and keep collecting
# digits if it is the first one.
ad5: clrl r6 # Initialize the exponent accumulator
cmpb r4,$'e # We allow both lower case e
cmpb r4,$'E # upper-case E
* Does the exponent have a sign?
ex1: movzbl (r2)+,r4 # Get next character
cmpb r4,$'+ # Positive sign?
cmpb r4,$'- # Negative sign?
bisb2 $1<esign,r3 # Indicate exponent is negative
ex2: movzbl (r2)+,r4 # Grab the next character
* Accumulate exponent digits in r6
ex3: cmpb r4,$'0 # A digit is within the range
cmpl r6,$214748364 # Exponent outrageously large already?
moval (r6)[r6],r6 # r6 *= 5
movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0'
jbr ex2 # Go 'round again
* Now get the final exponent and force it within a reasonable
* range so our scaling loops don't take forever for values
* that will ultimately cause overflow or underflow anyway.
* A tight check on over/underflow will be done by ldexp.
jbc $esign,r3,ex5 # Jump if exponent not negative
mnegl r6,r6 # If sign, negate exponent
ex5: addl2 r6,r5 # Add given exponent to calculated exponent
cmpl r5,$-100 # Absurdly small?
movl $-100,r5 # ... yes, force within limit
ex6: cmpl r5,$100 # Absurdly large?
movl $100,r5 # ... yes, force within bounds
* Our number has now been reduced to a mantissa and an exponent.
* The mantissa is a 63-bit positive binary integer in r0,r1,
* and the exponent is a signed power of 10 in r5. The msign
* bit in r3 will be on if the mantissa should ultimately be
* We now have to convert it to a standard format floating point
* number. This will be done by accumulating a binary exponent
* in r2, as we progressively get r5 closer to zero.
* Don't bother scaling if the mantissa is zero
movq r0,r0 # Mantissa zero?
clrl r2 # Initialize binary exponent
tstl r5 # Which way to scale?
jleq sd0 # Scale down if decimal exponent <= 0
* Scale up by "multiplying" r0,r1 by 10 as many times as necessary,
* Step 1: Shift r0,r1 right as necessary to ensure that no
* overflow can occur when multiplying.
su0: cmpl r1,$429496729 # Compare high word to (2**31)/5
jlss su1 # Jump out if guaranteed safe
ashq $-1,r0,r0 # Else shift right one bit
incl r2 # bump exponent to compensate
jbr su0 # and go back to test again.
* Step 2: Multiply r0,r1 by 5, by appropriate shifting and
* double-precision addition
su1: ashq $2,r0,r6 # (r6,r7) := (r0,r1) * 4
addl2 r6,r0 # Add low-order halves
adwc r7,r1 # and high-order halves
* Step 3: Increment the binary exponent to take care of the final
* factor of 2, and go back if we still need to scale more.
incl r2 # Increment the exponent
sobgtr r5,su0 # and back for more (maybe)
jbr cm0 # Merge to build final value
* Scale down. We must "divide" r0,r1 by 10 as many times
* Step 0: Right now, the condition codes reflect the state
* of r5. If it's zero, we are done.
sd0: jeql cm0 # If finished, build final number
* Step 1: Shift r0,r1 left until the high-order bit (not counting
* the sign bit) is nonzero, so that the division will preserve
* as much precision as possible.
tstl r1 # Is the entire high-order half zero?
jneq sd2 # ...no, go shift one bit at a time
ashq $30,r0,r0 # ...yes, shift left 30,
subl2 $30,r2 # decrement the exponent to compensate,
# and now it's known to be safe to shift
sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and
decl r2 # decrement the exponent to compensate
sd2: jbc $30,r1,sd1 # If the high-order bit is off, go shift
* Step 2: Divide the high-order part of (r0,r1) by 5,
* giving a quotient in r1 and a remainder in r7.
sd3: movl r1,r6 # Copy the high-order part
clrl r7 # Zero-extend to 64 bits
ediv $5,r6,r1,r7 # Divide (cannot overflow)
* Step 3: Divide the low-order part of (r0,r1) by 5,
* using the remainder from step 2 for rounding.
* Note that the result of this computation is unsigned,
* so we have to allow for the fact that an ordinary division
* by 5 could overflow. We make allowance by dividing by 10,
* multiplying the quotient by 2, and using the remainder
* to adjust the modified quotient.
addl3 $2,r0,r6 # Dividend is low part of (r0,r1) plus
adwc $0,r7 # 2 for rounding plus
# (2**32) * previous remainder
ediv $10,r6,r0,r6 # r0 := quotient, r6 := remainder.
addl2 r0,r0 # Make r0 result of dividing by 5
cmpl r6,$5 # If remainder is 5 or greater,
jlss sd4 # increment the adjustted quotient.
* Step 4: Increment the decimal exponent, decrement the binary
* exponent (to make the division by 5 into a division by 10),
* and back for another iteration.
sd4: decl r2 # Binary exponent
* We now have the following:
* r0: low-order half of a 64-bit integer
* r1: high-order half of the same 64-bit integer
* Our final result is the integer represented by (r0,r1)
* multiplied by 2 to the power contained in r2.
* We will transform (r0,r1) into a floating-point value,
* set the sign appropriately, and let ldexp do the
* Step 1: if the high-order bit (excluding the sign) of
* the high-order half (r1) is 1, then we have 63 bits of
* fraction, too many to convert easily. However, we also
* know we won't need them all, so we will just throw the
* low-order bit away (and adjust the exponent appropriately).
cm0: jbc $30,r1,cm1 # jump if no adjustment needed
ashq $-1,r0,r0 # lose the low-order bit
incl r2 # increase the exponent to compensate
* Step 2: split the 62-bit number in (r0,r1) into two
* 31-bit positive quantities
cm1: ashq $1,r0,r0 # put the high-order bits in r1
# and a 0 in the bottom of r0
rotl $-1,r0,r0 # right-justify the bits in r0
# moving the 0 from the ashq
* Step 3: convert both halves to floating point
cvtld r0,r6 # low-order part in r6-r7
cvtld r1,r0 # high-order part in r0-r1
* Step 4: multiply the high order part by 2**31 and combine them
muld2 two31,r0 # multiply
* Step 5: if appropriate, negate the floating value
jbc $msign,r3,cm2 # Jump if mantissa not signed
mnegd r0,r0 # If negative, make it so
* Step 6: call ldexp to complete the job
cm2: pushl r2 # Put exponent in parameter list
movd r0,-(sp) # and also mantissa
calls $3,_ldexp # go combine them
two31: .word 0x5000 # 2 ** 31
.word 0 # in floating-point
.word 0 # (so atof doesn't have to convert it)
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