x / 2 + 100 * (a + b) - 3 / (c + d) + e * e in assembly

Question

write an algorithm for: x / 2 + 100 * (a + b) - 3 / (c + d) + e * e knowing that: a, c - word, b, d - byte, e - doubleword, x - qword

    mov eax, dword [x]
    mov edx, dword [x + 4] ; edx:eax = x
    mov ebx, 2
    idiv ebx ; eax = edx:eax / ebx = x / 2
    mov ebx, eax ; save the result in ebx so we can do the other operations
    mov al, [b]
    cbw ; ax = b
    add ax, [a] ; ax = a + b
    mov dx, 100
    imul dx ; dx:ax = ax * dx = 100 * (a + b)
    push dx
    push ax
    pop eax ; 100 * (a + b)
    add ebx, eax ; ebx = x / 2 + 100 * (a + b)
    mov al, [d] ; al = d
    cbw ; ax = d
    add ax, word [c] ; ax = c + d
    mov cx, ax ; cx = c + d
    mov ax, 3
    cwd
    idiv cx ; ax = dx:ax / cx 
    cwd
    push dx
    push ax
    pop eax ; eax = 3 / (c + d)
    sub ebx, eax
    mov eax, ebx
    cdq ; edx:eax = x / 2 + 100 * (a + b) - 3 / (c + d)
    mov ebx, eax
    mov ecx, edx ; ecx:edx = x / 2 + 100 * (a + b) - 3 / (c + d)
    mov eax, [e]
    imul dword [e] ; edx:eax = e * e
    add eax, ebx
    adc edx, ecx
    mov dword [result + 0], eax 
    mov dword [result + 4], edx

did I make it unnecessarily complicated?

Answer 1

write an algorithm for: x / 2 + 100 * (a + b) - 3 / (c + d) + e * e

a, c - word,
b, d - byte,
e - doubleword,
x - qword

Because your biggest number is 64 bits (x is a qword), your final result will have to be 64 bits too!

Your first operation was to divide the qword in x by 2. You seem to expect that this result will fit in just a single dword because you've moved the quotient in the EBX register. You cannot make this assumption and worse the division could easily produce a divide exception if the quotient doesn't fit in 32 bits.
For the solution you should be aware that dividing by 2 is actually simply a shift to the right.

mov   ebx, [x]
mov   ebp, [x + 4] ; EBP:EBX is x
sar   ebp, 1
rcr   ebx, 1       ; EBP:EBX is x / 2

This means that you'll have to scale up the other calculations too in order to add them to EBP:EBX using:

add   ebx, ...
adc   ebp, ...

Because addition is associative, you can start by calculating the e * e part. You did not rearrange the expression and had to move around some more the registers in the end. Not a big deal, but nicer my way:

mov   eax, [e]
imul  eax
add   ebx, eax
adc   ebp, edx

Then comes 100 * (a + b):

movsx eax, word [a]
movsx edx, byte [b]
add   eax, edx       ; eax = a + b
mov   edx, 100
imul  edx            ; edx:eax = 100 * (a + b)
add   ebx, eax
adc   ebp, edx

I'll leave 3 / (c + d) to you...

... and finally the end will be:

sub   ebx, eax
sbb   ebp, edx
mov   [result + 0], ebx 
mov   [result + 4], ebp

---

did I make it unnecessarily complicated?

• It was a bit hard to read your program because you didn't insert some blank lines between the different operations.

• You don't need to write a size tag (byte, word, dword) if the register involved already implies the size. In mov dword [result + 0], eax the dword tag is redundant.

• Best have the comments in the program aligned above each other.

• Reread carefully to avoid typos like in:

    mov ecx, edx ; ecx:edx = x / 2 + 100 * (a + b) - 3 / (c + d)
  

  Should be ECX:EBX.

• To compute a square: once you've loaded the number in the register, you can multiply by that same register and not turn to the memory a second time like you did:

    mov   eax, [e]
    imul  eax        ; Don't write "imul dword [e]"