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8087 instruction set and examples. About the 8087 In the old days, you had to buy a -87 chip for numerical co-processing and hand install it. (Back in.

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Presentation on theme: "8087 instruction set and examples. About the 8087 In the old days, you had to buy a -87 chip for numerical co-processing and hand install it. (Back in."— Presentation transcript:

1 8087 instruction set and examples

2 About the 8087 In the old days, you had to buy a -87 chip for numerical co-processing and hand install it. (Back in the 8086, 286,386 days). With the 486 processor and later, the -87 coprocessor is integrated.

3 About the 8087 The 8087 uses a “stack” or chain of 8 registers to use for internal storage and data manipulation, as well as status and control words to set rounding control and indicate the results of operations. It has its own instruction set, instructions are recognizable because of the F- in front. (Like FIADD, FCOM, etc)

4 About the 8087 FWAIT… checks a control line to see if the 87 is still active. programmers used to have to use an FWAIT before and after a set of instructions for the -87 to make sure -86 or -87 storage operations had been completed. (Basically, to handle synchronization). FWAIT should not be necessary.

5 About the 8087: data transfer Data transfer: Instructionresult FLD sourceload a real into st(0) FST deststore real at dest (86 mem) FSTP deststore/pop st(0) FXCH {st(i)}exchange two regs (St(0), St(1)) or St(0) and the operand

6 About the 8087: data transfer FILD sourceint load FIST destint store FISTPdestint store/pop FBLD sourcebcd load (tbyte source) FBSTP destbcd store tbyte dest

7 About the 8087: arithmetic FADD sourceadd a real (mem) to st(0) FADD {St(1),ST} FADD st(i),st … or FADD ST,ST(i) FADDP St(i),STadd st to st(i) and pop st(0) FIADD {st,} sourceint add to st

8 About the 8087: arithmetic FSUB for real subtract has same formats as FADD FISUB {st,} intmem for int sub ALSO: FSUBR {st(1),st} FSUBR {st,} rmem FSUBRP st(i),st Etc and … FISUBR {st,} intmem Reverse subtract: subtract dest from source

9 About the 8087: arithmetic FMUL and FIMUL have same formats available FDIV and FIDIV have same formats Also available FDIVR, FDIVRP and FIDIVR

10 About the 8087: arithmetic Miscellaneous FSQRT {st} FABS {st} FCHS {ST}… change sign FLDZ {st}… load a zero

11 Control word The control word is a 16 bit word that works like a flag register on the 86 processor. It keeps information about zerodivide in bit 2 for example. Bits 3 and 4 are overflow and underflow respectively. Precision control is set in bits 8 and 9 and rounding is set in bits 10 and 11. Rounding control bit settings are: 00 round to nearest/even O1 round down 10 round up 11 chop/truncate

12 Status word Status word is also a 16 bit word value. condition bits are named c3, c2,c1 and c0. Their position in the status word, though is: C3 is bit 14 C2,c1,c0 are bits 10,9,8 respectively. If you are curious, the eight possible bit settings in bits (11,12,13) indicate which register in the chain is currently st(0)

13 Temp real (80 bits) Temp real has an f-p format. Bits are the significand in hidden bit format. Bits 64 to 78 are the biased exponent, bit 79 is the sign. You shouldn’t need to worry about these values on the stack.

14 Packed bcd (80 bits) A packed bcd is a tbyte. Bit 79 is the sign. Bits 72 through 78 are not used. Bits 0,1,2,3 store the 0 th (lsd) decimal digit. Bits 4,5,6,7 store the 1st. The 17 th (msd) digit is in bits 68,69,70,71. You’ll need to pad with zeros if there are fewer than 18 digits.

15 Int values 87 processor recognizes word, dword and qword signed int types, in the same manner as the 86 processor.

16 The stack Pushing and popping on the stack change the register who is currently the top. Pushing more than 8 times will push the last piece of data out and put St(0) at the most recently pushed item. It is the programmer’s responsibility to count stack push/pop operations. Whoever is on top of the stack, is referenced by St or ST(0). St(1) is next. And so on to St(7).

17 Int transfer FILD: load int. Type (word, dword, etc) is whatever the operand type is. St(0) is this new value. St(1) points to the previous value on top. FIST: copy St(0) value to memory, converting to a signed int (following rounding control settings in the control word) FISTP: same as above, but pop the stack as well.

18 BCD FBLD load a bcd (tbyte) onto the stack FBSTP store a tbyte bcd value into the memory operand, popping the stack as you go. Example: FBLD myval FBSTP myval

19 Exchanging/swapping on the stack FXCHG dest Swap stack top with operand. Example FXCHG St(3) ; swaps St(0) value with St(3) value

20 Int arithmetic FIADD, FIADD: add FISUB, FISUBR, : sub, or sub reversed. FIMUL FIDIV, FIDIVR Other: FPREM, FRNDINT – partial remainder, round to int FLDZ push a zero onto the stack. FLD1 push a 1

21 FPREM Takes implicit operands st,st(1) Does repeated subtract… leaves st>0,(possibly st==st(1)) or st==0. May need to repeat it, because it only reduces st by exp(2,64) If st>st(1) it needs to be repeated. FPREM sets bit C2 of the status word if it needs to be repeated, clears this bit if it completes operation.

22 operands Stack operands may be implicitly referenced. FIADD, FISUB, FIDIV, FIDIVR, FIMUL and FISUBR have only one form (example using add instruction): FIADD {ST,} intmem St(0) is implied dest for all of these.

23 comparison FCOM ;no operands compares st,st(1) FCOM St(i); one operand compares st with st(i) FCOMP (compare and pop) is the same. FCOMPP - only allows implicit operands St,st(1) (compare then pop twice) FTST – compares St with 0. These all use condition codes described above and make the settings in the next slide.

24 Condition codes C3c0 00st>source 01st

25 Getting status and control words FSTSW intmem ; copy status word to 16 bit mem location for examination. FLDCW intmem; load control word (to set rounding, for example, from 16bit int mem) FSTCW intmem; copy control word to int mem

26 Some -87 examples

27 16-bit vs 32-bit? Almost all the examples here were done in 32 bit. But the coprocessor works in 16 bit as well.

28 excerpt from a 16-bit program & ouptut value word 1234.code main PROC mov mov ds,ax mov ax, value call writedec fild value fiadd value fistp value mov ax,value call crlf call writedec C:\MASM615>coprocessor C:\MASM615>

29 adding up an array (example from notes) include irvine16.inc.data array dword 12, 33, 44,55,77,88,99,101,202,9999,111 sum dword ?.code main proc mov mov ds,ax xor bx,bx fldz fist sum mov eax,sum;;; call writeint;print zero call crlf mov cx,10 top: mov eax,array[bx];;get value call writeint;;;print it call crlf fiadd array[bx];;;;add to st(0) add bx,4 loop top fistp sum;;;;when done pop to dword int mov eax,sum call writeInt;;;;print it mov ax,4c00h int 21h main endp end main

30 adding up an array (example from notes)

31 Getting sqrt call readint fstcw control;;;store current control word or control,0800h;set bit 11 clear bit 10 to round up fldcw control;;load new control word mov num,eax fild num fsqrt fistp sqr fwait mov edx, offset prompt2 call writestring mov eax,sqr call writeint

32 Rounding set to round up c:\Masm615>primes enter an integer125 sqrt of integer+12

33 Add code to check for prime and print divisors for composites mov eax,num call crlf mov ebx,2;;;first divisor top: xor edx,edx push eax div ebx;;divide mov divi,ebx cmp edx,0 je notprime inc ebx cmp ebx,sqr jg prime pop eax jmp top notprime: call writedec call crlf mov eax,divi call writedec call crlf mov edx,offset f call writestring prime: mov edx,offset t call writestring

34 Output from primes enter an integer not a prime c:\Masm615>primes enter an integer not a prime c:\Masm615>primes enter an integer not a prime c:\Masm615>primes enter an integer prime c:\Masm615>

35 factorials call readint mov num,eax call crlf fld1 ;;load a 1 for subtacting and comparing..this will be st(2) fld1 ;;prod value initialized in st(1) fild num;;;multiplier... need to count down and multiply st by st(1) theloop: ftst ;;is st==0? fstsw status and status, 4100h;check bits c3 and c0...c3=0 c0=1 means st

36 factorials c:\Masm615>factorials enter an integer6 factorial is 720 c:\Masm615>factorials enter an integer7 factorial is 5040 c:\Masm615>

37 Fibonacci values mov edx, offset prompt call crlf call writestring call readint call crlf fld1 ;;load a 1 for subtacting and comparing..this will be st(2) fld1 ;;prod value initialized in st(1) top: cmp eax,0 je done fadd st(1),st fsubr st,st(1);;;cute huhn?st=st(1)-st dec eax jmp top done: fistp dummy fistp answer mov edx,offset p2 call writestring call crlf fwait mov eax,answer call writedec

38 Fibonacci c:\Masm615>fibs enter an integer4 fib is 8 c:\Masm615>fibs enter an integer6 fib is 21 c:\Masm615>fibs enter an integer7 fib is 34 c:\Masm615>

39 Mimicking real io You can output reals by outputting first the integer part, subtracting this from the original value, then repeatedly multiplying the fraction by ten, (subtracting this off from the remainder) and outputting this sequence of fractional digits. This is NOT an IEEE f-p conversion routine!

40 Rounding control & the int part fstcw control or control,0C00h ;;;chop or truncate fldcw control fild ten ;;ten is in st(1) fild num ;;num is in st(0) fsqrt ;;sqrt in st fist intsqr ;;;store chopped result, don’t pop call crlf mov edx,offset message call writestring fwait mov ax,intsqr ;;write the int part call writedec

41 realio mov edx,offset dot;;decimal point call writestring fisub intsqr ;;subtract from sqrt the int part leaving fractional part ;now loop & store 5 decimal places mov edi, offset decimals mov ecx, 5 up: fmul st,st(1) ;;multiply by 10 to shift dec point fist digit ;store truncated int fisub digit ;;subtract it off of the total fwait mov ax,digit add al,48 mov byte ptr [edi],al ;;store this digit inc edi loop up

42 Run of realio c:\Masm615>realio enter an integer: 121 sqrt of integer c:\Masm615>realio enter an integer: 123 sqrt of integer c:\Masm615>realio enter an integer: 143 sqrt of integer c:\Masm615>


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