Question

For exmaple:

How to represent the following x86 in SSA form:

xor  eax, eax
inc  ax

By introducing some pseudo functions, I come up with:

eax@1 = eax@0 ^ eax@0
ax@1 = LOWORD(eax@1)
al@1 = LOBYTE(ax@1)
ah@1 = HIBYTE(ax@1)
hax@1 = HIWORD(eax@1)

ax@2 = ax@1 + 1
eax@2 = MAKEDWORD(ax@2, HIWORD(eax@1))
al@2 = LOBYTE(ax@2)
ah@2 = HIBYTE(ax@2)

But I think it’s too much verbose

Answer 1

Using your notation:

1. eax@0 = … whatever it was before here …
2. eax@1 = 0
3. ax@2 = ax@1 + 1

Because eax contains ax, there’s an implicit step in between 2 and 3

1. eax@0 = …
2. eax@1 = 0
3. ax@1 = 0 (because ax cannot be non-zero if eax is zero)
4. ax@2 = ax@1 + 1

Step 2 because any number xor’ed with itself is 0… eax@0 is dead at that point, and thus eax@1 can be renamed (using ebx as renaming so it’s readable; obviously you would use a virtual register, not a real one):

1. — deleted, eax no longer relevant
2. ebx@0 = 0
3. bx@0 = 0
4. bx@1 = bx@0 + 1

You could then note that because step 3 is a constant function, so is step 4 (adding a constant to a constant) and compress the two together (i.e. constant folding)

1. — deleted, eax no longer relevant
2. ebx@0 = 0
3. bx@0 = 1

If the upper 16 bits of ebx don’t dominate anything below this, you could also delete step 2.