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ORCA-C/Expression.asm

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Change all text/source files to LF line endings.
2017-10-21 23:40:19 +00:00
mcopy exp.macros
****************************************************************
*
* function lshr(x,y: longint): longint;
*
* Inputs:
* num1 - number to shift
* num2 - # bits to shift by
*
* Outputs:
* A - result
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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lshr start exp
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subroutine (4:num1,4:num2),0
lda num2+2 if num2 < 0 then
bpl lb2
cmp #$FFFF shift left
bne zero
ldx num2
cpx #-34
blt zero
lb1 asl num1
rol num1+2
inx
bne lb1
bra lb4
zero stz num1 (result is zero)
stz num1+2
bra lb4
lb2 bne zero else shift right
ldx num2
beq lb4
cpx #33
bge zero
lb3 lsr num1+2
ror num1
dex
bne lb3
lb4 lda 0 fix stack and return
sta num2
lda 2
sta num2+2
return 4:num1
end
****************************************************************
*
* function udiv(x,y: longint): longint;
*
* Inputs:
* num1 - numerator
* num2 - denominator
*
* Outputs:
* ans - result
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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udiv start exp
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ans equ 0 answer
rem equ 4 remainder
subroutine (4:num1,4:num2),8
;
; Initialize
;
stz rem rem = 0
stz rem+2
move4 num1,ans ans = num1
lda num2 check for division by zero
ora num2+2
beq dv9
lda num2+2 do 16 bit divides separately
ora ans+2
beq dv5
;
; 32 bit divide
;
ldy #32 32 bits to go
dv3 asl ans roll up the next number
rol ans+2
rol ans+4
rol ans+6
sec subtract for this digit
lda ans+4
Fix bug in unsigned 32-bit division and remainder routines, which could cause mis-evaluation of constant expressions. The following example shows cases that were mis-evaluated: /* Should print "3 10000" */ #include <stdio.h> int main(void) { printf("%lu %lu\n", 100000ul / 30000ul, 100000ul % 30000ul); }
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sbc num2
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tax
lda ans+6
sbc num2+2
bcc dv4 branch if minus
stx ans+4 turn the bit on
sta ans+6
inc ans
dv4 dey next bit
bne dv3
bra dv9 go do the sign
;
; 16 bit divide
;
dv5 lda #0 initialize the remainder
ldy #16 16 bits to go
dv6 asl ans roll up the next number
rol a
sec subtract the digit
sbc num2
bcs dv7
adc num2 digit is 0
dey
bne dv6
bra dv8
dv7 inc ans digit is 1
dey
bne dv6
dv8 sta ans+4 save the remainder
;
; Return the result
;
dv9 return 4:ans move answer
end
****************************************************************
*
* function uge(x,y: longint): cboolean;
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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uge start exp
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result equ 0
subroutine (4:x,4:y),4
stz result
stz result+2
lda x+2
cmp y+2
bne lb1
lda x
cmp y
lb1 blt lb2
Fix to make statically-evaluated unsigned comparisons evaluate to 1 if true. Previously, they evaluated to -1. The following example shows the problem: #include <stdio.h> int main(void) { printf("%i\n", 0U <= 5U); }
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inc result
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lb2 return 2:result
end
****************************************************************
*
* function ugt(x,y: longint): cboolean;
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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ugt start exp
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result equ 0
subroutine (4:x,4:y),4
stz result
stz result+2
lda x+2
cmp y+2
bne lb1
lda x
cmp y
lb1 ble lb2
Fix to make statically-evaluated unsigned comparisons evaluate to 1 if true. Previously, they evaluated to -1. The following example shows the problem: #include <stdio.h> int main(void) { printf("%i\n", 0U <= 5U); }
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inc result
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lb2 return 2:result
end
****************************************************************
*
* function ule(x,y: longint): cboolean;
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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ule start exp
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result equ 0
subroutine (4:x,4:y),4
stz result
stz result+2
lda x+2
cmp y+2
bne lb1
lda x
cmp y
lb1 bgt lb2
Fix to make statically-evaluated unsigned comparisons evaluate to 1 if true. Previously, they evaluated to -1. The following example shows the problem: #include <stdio.h> int main(void) { printf("%i\n", 0U <= 5U); }
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inc result
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lb2 return 2:result
end
****************************************************************
*
* function ult(x,y: longint): cboolean;
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
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ult start exp
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result equ 0
subroutine (4:x,4:y),4
stz result
stz result+2
lda x+2
cmp y+2
bne lb1
lda x
cmp y
lb1 bge lb2
Fix to make statically-evaluated unsigned comparisons evaluate to 1 if true. Previously, they evaluated to -1. The following example shows the problem: #include <stdio.h> int main(void) { printf("%i\n", 0U <= 5U); }
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inc result
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lb2 return 2:result
end
****************************************************************
*
* function umod(x,y: longint): longint;
*
* Inputs:
* num1 - numerator
* num2 - denominator
*
* Outputs:
* ans+4 - result
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
2017-06-19 03:07:32 +00:00
umod start exp
Change all text/source files to LF line endings.
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ans equ 0 answer
rem equ 4 remainder
subroutine (4:num1,4:num2),8
;
; Initialize
;
stz rem rem = 0
stz rem+2
move4 num1,ans ans = num1
lda num2 check for division by zero
ora num2+2
beq dv9
lda num2+2 do 16 bit divides separately
ora ans+2
beq dv5
;
; 32 bit divide
;
ldy #32 32 bits to go
dv3 asl ans roll up the next number
rol ans+2
rol ans+4
rol ans+6
sec subtract for this digit
lda ans+4
Fix bug in unsigned 32-bit division and remainder routines, which could cause mis-evaluation of constant expressions. The following example shows cases that were mis-evaluated: /* Should print "3 10000" */ #include <stdio.h> int main(void) { printf("%lu %lu\n", 100000ul / 30000ul, 100000ul % 30000ul); }
2016-12-19 04:52:53 +00:00
sbc num2
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tax
lda ans+6
sbc num2+2
bcc dv4 branch if minus
stx ans+4 turn the bit on
sta ans+6
inc ans
dv4 dey next bit
bne dv3
bra dv9 go do the sign
;
; 16 bit divide
;
dv5 lda #0 initialize the remainder
ldy #16 16 bits to go
dv6 asl ans roll up the next number
rol a
sec subtract the digit
sbc num2
bcs dv7
adc num2 digit is 0
dey
bne dv6
bra dv8
dv7 inc ans digit is 1
dey
bne dv6
dv8 sta ans+4 save the remainder
;
; Return the result
;
dv9 return 4:ans+4 move answer
end
****************************************************************
*
* function umul(x,y: longint): longint;
*
* Inputs:
* num2,num1 - operands
*
* Outputs:
* ans - result
*
****************************************************************
*
Move some code from the blank segment to named load segments. This frees up some space in the blank segment for more static data.
2017-06-19 03:07:32 +00:00
umul start exp
Change all text/source files to LF line endings.
2017-10-21 23:40:19 +00:00
ans equ 0 answer
subroutine (4:num1,4:num2),8
;
; Initialize the sign and split on precision.
;
stz ans+4 set up the multiplier
stz ans+6
lda num1
sta ans
lda num1+2
sta ans+2
beq ml3 branch if the multiplier is 16 bit
;
; Do a 32 bit by 32 bit multiply.
;
ldy #32 32 bit multiply
jsr ml1
brl ml7
ml1 lda ans SYSS1*SYSS1+2+SYSS1+2 -> SYSS1,SYSS1+2
lsr a
bcc ml2
clc add multiplicand to the partial product
lda ans+4
adc num2
sta ans+4
lda ans+6
adc num2+2
sta ans+6
ml2 ror ans+6 shift the interem result
ror ans+4
ror ans+2
ror ans
dey loop til done
bne ml1
rts
;
; Do and 16 bit by 32 bit multiply.
;
ml3 lda num2+2 branch if 16x16 is possible
beq ml4
ldy #16 set up for 16 bits
jsr ml1 do the multiply
lda ans+2 move the answer
sta ans
lda ans+4
sta ans+2
bra ml7
;
; Do a 16 bit by 16 bit multiply.
;
ml4 ldy #16 set the 16 bit counter
ldx ans move the low word
stx ans+2
ml5 lsr ans+2 test the bit
bcc ml6 branch if the bit is off
clc
adc num2
ml6 ror a shift the answer
ror ans
dey loop
bne ml5
sta ans+2 save the high word
;
; Return the result.
;
ml7 return 4:ans fix the stack
end
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