CAN EMU8086 DO 64 BIT APPLICATIONS 64 BITS
CAN EMU8086 DO 64 BIT APPLICATIONS CODE

With x86 - some instruction can surprise you, for example the code above modified to: mov ax, description, including which flags it does affect. The lesson to learn here is, that you should often check instruction reference guide for detailed ins. The add si,2 will modify the CF, so the next sbb does not 'continue' in the subtraction. You can do the same without using 4 16b registers to hold the single 64b value, like: mov ax,ДЄfter looking at that one, you may get a great idea, how about: mov ax,ДЄdd si,2 adjust rather base pointer then changing all those offsetsДoes it work? Not any more. so let's add some more info at least, to make you work it out a bit. You can store CF somewhere if you wish, the CPU has instructions to detect/store/set value of CF, but with numbers once you calculate 15 - 8 is 7, the 7 is final, you don't need to remember you did borrow the '1' to subtract 8 from 5.ДЄnd the result address is again easy to calculate relatively to num1 address, so you can just store the result back into memory mov ,ax 0x0150 + 0x20 = 0x0170ДЄnd I basically gave you whole solution :/. the result modified by CF is already stored in bx, the old CF is lost, CF contains new 'borrow' of bx - old_borrow. happening while subtracting the LSW to propagate into upper 16 bitsДЄbout your question in comments. but this time CF will affect the result, to make any 'borrowing' Sbb bx, continuing by subtracting adjoining 16 bits here dx:cx:bx:ax concatenated into single 64b number contains the number1Д«ut the number2 can be addressed by base address of number1 too, so you can also continue by subtraction: sub ax, subtraction of LSW, ignoring CF (borrow) But actually you can use relative addressing in 16b mode, so this is possible: mov si,0x0150 set si to contain address of input number1 Now if you would want to put all addresses of all words (3 * 4 words of memory will be used, num1, num2 and result, all of them 4 words long), you would run out of registers quickly. Mov ax, usually not practical, but possible you can even use absolute addressing like: Mov ax, load least significant 16b word (LSW) of it into ax

If you want to load the actual value from memory, you have to dereference that memory address, like for example: mov si,0x0150 set si to contain address of input number1 That description sounds like by default a memory unit dword (32 bits) is used, which in 16b real mode is not very sensible default data size.Дў) mov ax,0x0150 does not load the value from memory, it loads that constant 0x0150 into ax. That's not very exact task description, the 64b integer is in memory at locations from 0x0150 to 0x0157 (8 bytes in total). The 64-bit integer in memory locations 0x0150 and 0x154 SO ANY push in the right direction will hopefully get me on my way! This is the first of 12 similar programs I have to write. Is that hex format okay? or does it need to be something like 0174h?Д®ven if this is somehow correct, do I need a return statement or a header or anything to make this actually compile? (I'm using NASM and DosBox) Then I subtracted the registers and put them into the memory locations. I stored each half of the 64 bit integer into the registers. Maybe I'm not that far off, but it feels like it. Does it matter which GPIO we use? This is my example, so maybe you can see how I'm thinking. And I know we subtract the least significant pieces first.

CAN EMU8086 DO 64 BIT APPLICATIONS 64 BITS

I understand the logic behind separating it into smaller pieces since we can't fit 64 bits into the registers. Store the result in memory location 0x0170 and 0x0174. Write a program called SUB64 to subtract the 64-bit integer in memory locations 0x0150 and 0x0154 from the 64-bit integer in 0x0160 and 0x0164.

Can emu8086 do 64 bit applications