tiny optimisation to gpt entry size check

boot0.s

@@ -126,13 +126,31 @@ STAGE2_GPT_ENTRY_ADDR  equ 0x1a
   jz .panic
   and eax, 127 ; Test size is a multiple of 128
   jnz .panic
+  ; Use the (n & (n - 1)) == 0 trick to test if the entry size is a power of 2. Since we already
+  ; know it's a nonzero multiple of 128, if size is a power of 2 then size = 128 * 2^n holds
+  ; (proof below because it felt correct to me but it didn't feel obvious enough to use without
+  ; proving). Therefore we don't need to bother dividing by 128 first (shr 7), which saves a couple
+  ; of bytes.
+  ; - Let s = 128 * a, s = 2^b for integers a >= 1, b >= 0
+  ; - Lemma: b >= 7
+  ;   - Assume b < 7
+  ;   - s = 2^b
+  ;   - s < 2^7
+  ;   - s < 128
+  ;   - s = 128 * a, a >= 1
+  ;   - s/128 >= 1
+  ;   - s >= 128
+  ;   - Contradiction: s < 128 and s >= 128 cannot both hold
+  ; - 2^7 * a = 2^b
+  ; - a = 2^(b-7)
+  ; - s = 128 * 2^(b-7)
+  ; - Therefore s = 128 * 2^n where n = b - 7.
+  ; - n >= 0 because b >= 7.
   mov eax, ebx
-  shr eax, 7   ; Divide by 128
-  mov ecx, eax
-  dec eax
-  and eax, ecx ; Test size/128 is a power of 2. popcnt gives an exception for some reason...
+  mov ecx, ebx
+  dec ecx
+  and ecx, eax
   jnz .panic
-  mov eax, ebx
 
   ; Find the "sector stride", which is the number of sectors we increment by each time we want to
   ; load a new entry.