marziehk
دوشنبه 21 دی 1394, 22:43 عصر
سلام کسی می دونه چرا این برنامه تو خط 64 خطا می گیره؟
; A small program that calculates and prints terms of the Fibonacci series
; fibo.asm
; assemble using nasm:
; nasm -o fibo.com -f bin fibo.asm
;
;************************************************* ***************************
; Alterable Constant
;************************************************* ***************************
; You can adjust this upward but the upper limit is around 150000 terms.
; the limitation is due to the fact that we can only address 64K of memory
; in a DOS com file, and the program is about 211 bytes long and the
; address space starts at 100h. So that leaves roughly 65000 bytes to
; be shared by the two terms (num1 and num2 at the end of this file). Since
; they're of equal size, that's about 32500 bytes each, and the 150000th
; term of the Fibonacci sequence is 31349 digits long.
;
maxTerms equ 15000 ; number of terms of the series to calculate
;************************************************* ***************************
; Number digits to use. This is based on a little bit of tricky math.
; One way to calculate F(n) (i.e. the nth term of the Fibonacci seeries)
; is to use the equation int(phi^n/sqrt(5)) where ^ means exponentiation
; and phi = (1 + sqrt(5))/2, the "golden number" which is a constant about
; equal to 1.618. To get the number of decimal digits, we just take the
; base ten log of this number. We can very easily see how to get the
; base phi log of F(n) -- it's just n*lp(phi)+lp(sqrt(5)), where lp means
; a base phi log. To get the base ten log of this we just divide by the
; base ten log of phi. If we work through all that math, we get:
;
; digits = terms * log(phi) + log(sqrt(5))/log(phi)
;
; the constants below are slightly high to assure that we always have
; enough room. As mentioned above the 150000th term has 31349 digits,
; but this formula gives 31351. Not too much waste there, but I'd be
; a little concerned about the stack!
;
digits equ (maxTerms*209+1673)/1000
; this is just the number of digits for the term counter
cntDigits equ 6 ; number of digits for counter
org 100h ; this is a DOS com file
;************************************************* ***************************
;************************************************* ***************************
main:
; initializes the two numbers and the counter. Note that this assumes
; that the counter and num1 and num2 areas are contiguous!
;
mov ax,'00' ; initialize to all ASCII zeroes
mov di,counter ; including the counter
mov cx,digits+cntDigits/2 ; two bytes at a time
cld ; initialize from low to high memory
rep stosw ; write the data
inc ax ; make sure ASCII zero is in al
mov [num1 + digits - 1],al ; last digit is one
mov [num2 + digits - 1],al ;
mov [counter + cntDigits - 1],al
jmp .bottom ; done with initialization, so begin
.top
; add num1 to num2
mov di,num1+digits-1
mov si,num2+digits-1
mov cx,digits ;
call AddNumbers ; num2 += num1
mov bp,num2 ;
call PrintLine ;
dec dword [term] ; decrement loop counter
jz .done ;
; add num2 to num1
mov di,num2+digits-1
mov si,num1+digits-1
mov cx,digits ;
call AddNumbers ; num1 += num2
.bottom
mov bp,num1 ;
call PrintLine ;
dec dword [term] ; decrement loop counter
jnz .top ;
.done
call CRLF ; finish off with CRLF
mov ax,4c00h ; terminate
int 21h ;
;************************************************* ***************************
;
; PrintLine
; prints a single line of output containing one term of the
; Fibonacci sequence. The first few lines look like this:
;
; Fibonacci(1): 1
; Fibonacci(2): 1
; Fibonacci(3): 2
; Fibonacci(4): 3
;
; INPUT: ds:bp ==> number string, cx = max string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx, di
;
;************************************************* ***************************
PrintLine:
mov dx,eol ; print combined CRLF and msg1
mov cx,msg1len+eollen ;
call PrintString ;
mov di,counter ; print counter
mov cx,cntDigits ;
call PrintNumericString
call IncrementCount ; also increment the counter
mov dx,msg2 ; print msg2
mov cx,msg2len ;
call PrintString ;
mov di,bp ; recall address of number
mov cx,digits ;
; deliberately fall through to PrintNumericString
;************************************************* ***************************
;
; PrintNumericString
; prints the numeric string at DS:DI, suppressing leading zeroes
; max length is CX
;
; INPUT: ds:di ==> number string, cx = max string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx, di
;
;************************************************* ***************************
PrintNumericString:
; first scan for the first non-zero byte
mov al,'0' ; look for ASCII zero
cld ; scan from MSD to LSD
repe scasb ;
mov dx,di ; points to one byte after
dec dx ; back up one character
inc cx ;
; deliberately fall through to PrintString
;************************************************* ***************************
;
; PrintString
; prints the string at DS:DX with length CX to stdout
;
; INPUT: ds:dx ==> string, cx = string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx
;
;************************************************* ***************************
PrintString:
mov bx, 1 ; write to stdout
mov ah, 040h ; write to file handle
int 21h ; ignore return value
ret ;
;************************************************* ***************************
;
; AddNumbers
; add number 2 at ds:si to number 1 at es:di of width cx
;
;
; INPUT: es:di ==> number1, ds:si ==> number2, cx= max width
; OUTPUT: CF set on overflow
; DESTROYED: ax, si, di
;
;************************************************* ***************************
AddNumbers:
std ; go from LSB to MSB
clc ;
pushf ; save carry flag
.top
mov ax,0f0fh ; convert from ASCII BCD to BCD
and al,[si] ; get next digit of number2 in al
and ah,[di] ; get next digit of number1 in ah
popf ; recall carry flag
adc al,ah ; add these digits
aaa ; convert to BCD
pushf ;
add al,'0' ; convert back to ASCII BCD digit
stosb ; save it and increment both counters
dec si ;
loop .top ; keep going until we've got them all
popf ; recall carry flag
ret ;
;************************************************* ***************************
;
; IncrementCount
; increments a multidigit term counter by one
;
; INPUT: none
; OUTPUT: CF set on overflow
; DESTROYED: ax, cx, di
;
;************************************************* ***************************
IncrementCount:
mov cx,cntDigits ;
mov di,counter+cntDigits-1
std ; go from LSB to MSB
stc ; this is our increment
pushf ; save carry flag
.top
mov ax,000fh ; convert from ASCII BCD to BCD
and al,[di] ; get next digit of counter in al
popf ; recall carry flag
adc al,ah ; add these digits
aaa ; convert to BCD
pushf ;
add al,'0' ; convert back to ASCII BCD digit
stosb ; save and increment counter
loop .top ;
popf ; recall carry flag
ret ;
;************************************************* ***************************
;
; CRLF
; prints carriage return, line feed pair to stdout
;
; INPUT: none
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx
;
;************************************************* ***************************
CRLF: mov dx,eol ;
mov cx,eollen ;
jmp PrintString ;
;************************************************* ***************************
; static data
;************************************************* ***************************
eol db 13,10 ; DOS-style end of line
eollen equ $ - eol
msg1 db 'Fibonacci(' ;
msg1len equ $ - msg1
msg2 db '): ' ;
msg2len equ $ - msg2
;************************************************* ***************************
; initialized data
;************************************************* ***************************
term dd maxTerms ;
;************************************************* ***************************
; unallocated data
;
; A better way to do this would be to actually ask for a memory
; allocation and use that memory space, but this is a DOS COM file
; and so we are given the entire 64K of space. Technically, this
; could fail since we *might* be running on a machine which doesn't
; have 64K free. If you're running on such a memory poor machine,
; my advice would be to not run this program.
;
;************************************************* ***************************
; static data
counter: ;
num1 equ counter+cntDigits ;
num2 equ num1+digits ;
; A small program that calculates and prints terms of the Fibonacci series
; fibo.asm
; assemble using nasm:
; nasm -o fibo.com -f bin fibo.asm
;
;************************************************* ***************************
; Alterable Constant
;************************************************* ***************************
; You can adjust this upward but the upper limit is around 150000 terms.
; the limitation is due to the fact that we can only address 64K of memory
; in a DOS com file, and the program is about 211 bytes long and the
; address space starts at 100h. So that leaves roughly 65000 bytes to
; be shared by the two terms (num1 and num2 at the end of this file). Since
; they're of equal size, that's about 32500 bytes each, and the 150000th
; term of the Fibonacci sequence is 31349 digits long.
;
maxTerms equ 15000 ; number of terms of the series to calculate
;************************************************* ***************************
; Number digits to use. This is based on a little bit of tricky math.
; One way to calculate F(n) (i.e. the nth term of the Fibonacci seeries)
; is to use the equation int(phi^n/sqrt(5)) where ^ means exponentiation
; and phi = (1 + sqrt(5))/2, the "golden number" which is a constant about
; equal to 1.618. To get the number of decimal digits, we just take the
; base ten log of this number. We can very easily see how to get the
; base phi log of F(n) -- it's just n*lp(phi)+lp(sqrt(5)), where lp means
; a base phi log. To get the base ten log of this we just divide by the
; base ten log of phi. If we work through all that math, we get:
;
; digits = terms * log(phi) + log(sqrt(5))/log(phi)
;
; the constants below are slightly high to assure that we always have
; enough room. As mentioned above the 150000th term has 31349 digits,
; but this formula gives 31351. Not too much waste there, but I'd be
; a little concerned about the stack!
;
digits equ (maxTerms*209+1673)/1000
; this is just the number of digits for the term counter
cntDigits equ 6 ; number of digits for counter
org 100h ; this is a DOS com file
;************************************************* ***************************
;************************************************* ***************************
main:
; initializes the two numbers and the counter. Note that this assumes
; that the counter and num1 and num2 areas are contiguous!
;
mov ax,'00' ; initialize to all ASCII zeroes
mov di,counter ; including the counter
mov cx,digits+cntDigits/2 ; two bytes at a time
cld ; initialize from low to high memory
rep stosw ; write the data
inc ax ; make sure ASCII zero is in al
mov [num1 + digits - 1],al ; last digit is one
mov [num2 + digits - 1],al ;
mov [counter + cntDigits - 1],al
jmp .bottom ; done with initialization, so begin
.top
; add num1 to num2
mov di,num1+digits-1
mov si,num2+digits-1
mov cx,digits ;
call AddNumbers ; num2 += num1
mov bp,num2 ;
call PrintLine ;
dec dword [term] ; decrement loop counter
jz .done ;
; add num2 to num1
mov di,num2+digits-1
mov si,num1+digits-1
mov cx,digits ;
call AddNumbers ; num1 += num2
.bottom
mov bp,num1 ;
call PrintLine ;
dec dword [term] ; decrement loop counter
jnz .top ;
.done
call CRLF ; finish off with CRLF
mov ax,4c00h ; terminate
int 21h ;
;************************************************* ***************************
;
; PrintLine
; prints a single line of output containing one term of the
; Fibonacci sequence. The first few lines look like this:
;
; Fibonacci(1): 1
; Fibonacci(2): 1
; Fibonacci(3): 2
; Fibonacci(4): 3
;
; INPUT: ds:bp ==> number string, cx = max string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx, di
;
;************************************************* ***************************
PrintLine:
mov dx,eol ; print combined CRLF and msg1
mov cx,msg1len+eollen ;
call PrintString ;
mov di,counter ; print counter
mov cx,cntDigits ;
call PrintNumericString
call IncrementCount ; also increment the counter
mov dx,msg2 ; print msg2
mov cx,msg2len ;
call PrintString ;
mov di,bp ; recall address of number
mov cx,digits ;
; deliberately fall through to PrintNumericString
;************************************************* ***************************
;
; PrintNumericString
; prints the numeric string at DS:DI, suppressing leading zeroes
; max length is CX
;
; INPUT: ds:di ==> number string, cx = max string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx, di
;
;************************************************* ***************************
PrintNumericString:
; first scan for the first non-zero byte
mov al,'0' ; look for ASCII zero
cld ; scan from MSD to LSD
repe scasb ;
mov dx,di ; points to one byte after
dec dx ; back up one character
inc cx ;
; deliberately fall through to PrintString
;************************************************* ***************************
;
; PrintString
; prints the string at DS:DX with length CX to stdout
;
; INPUT: ds:dx ==> string, cx = string length
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx
;
;************************************************* ***************************
PrintString:
mov bx, 1 ; write to stdout
mov ah, 040h ; write to file handle
int 21h ; ignore return value
ret ;
;************************************************* ***************************
;
; AddNumbers
; add number 2 at ds:si to number 1 at es:di of width cx
;
;
; INPUT: es:di ==> number1, ds:si ==> number2, cx= max width
; OUTPUT: CF set on overflow
; DESTROYED: ax, si, di
;
;************************************************* ***************************
AddNumbers:
std ; go from LSB to MSB
clc ;
pushf ; save carry flag
.top
mov ax,0f0fh ; convert from ASCII BCD to BCD
and al,[si] ; get next digit of number2 in al
and ah,[di] ; get next digit of number1 in ah
popf ; recall carry flag
adc al,ah ; add these digits
aaa ; convert to BCD
pushf ;
add al,'0' ; convert back to ASCII BCD digit
stosb ; save it and increment both counters
dec si ;
loop .top ; keep going until we've got them all
popf ; recall carry flag
ret ;
;************************************************* ***************************
;
; IncrementCount
; increments a multidigit term counter by one
;
; INPUT: none
; OUTPUT: CF set on overflow
; DESTROYED: ax, cx, di
;
;************************************************* ***************************
IncrementCount:
mov cx,cntDigits ;
mov di,counter+cntDigits-1
std ; go from LSB to MSB
stc ; this is our increment
pushf ; save carry flag
.top
mov ax,000fh ; convert from ASCII BCD to BCD
and al,[di] ; get next digit of counter in al
popf ; recall carry flag
adc al,ah ; add these digits
aaa ; convert to BCD
pushf ;
add al,'0' ; convert back to ASCII BCD digit
stosb ; save and increment counter
loop .top ;
popf ; recall carry flag
ret ;
;************************************************* ***************************
;
; CRLF
; prints carriage return, line feed pair to stdout
;
; INPUT: none
; OUTPUT: CF set on error, AX = error code if carry set
; DESTROYED: ax, bx, cx, dx
;
;************************************************* ***************************
CRLF: mov dx,eol ;
mov cx,eollen ;
jmp PrintString ;
;************************************************* ***************************
; static data
;************************************************* ***************************
eol db 13,10 ; DOS-style end of line
eollen equ $ - eol
msg1 db 'Fibonacci(' ;
msg1len equ $ - msg1
msg2 db '): ' ;
msg2len equ $ - msg2
;************************************************* ***************************
; initialized data
;************************************************* ***************************
term dd maxTerms ;
;************************************************* ***************************
; unallocated data
;
; A better way to do this would be to actually ask for a memory
; allocation and use that memory space, but this is a DOS COM file
; and so we are given the entire 64K of space. Technically, this
; could fail since we *might* be running on a machine which doesn't
; have 64K free. If you're running on such a memory poor machine,
; my advice would be to not run this program.
;
;************************************************* ***************************
; static data
counter: ;
num1 equ counter+cntDigits ;
num2 equ num1+digits ;