Two idioms, zero new instructions
This lesson stacks two things you have already met. An enum is int-sized and adds nothing of its own to the generated code, from lesson 5, and a dense switch dispatches through a jump table, from lesson 1. Combine them into a switch over an enum-typed argument with consecutive values and the two never interfere. The enum is only renaming integers, so the dispatch matches what a switch on a plain int would emit.
Take paint(Brush b), which switches over an eight-value Brush enum (BRUSH_PEN, BRUSH_FILL, and so on) and returns a pixel cost per tool:
cmplwi r3, 7 # b arrives as a 4-byte int; bounds-check 0..7
bgt- .default
lis r4, table@ha
slwi r0, r3, 2 # b * 4 -> table index
addi r3, r4, table@lo
lwzx r0, r3, r0
mtctr r0
bctr # jump straight to the case for this brush
.case0: li r3, 10 blr # each arm is one li/blr...
.case1: li r3, 25 blr # ...the enum names left no trace
That cmplwi/lwzx/mtctr/bctr shape is lesson 1's jump table, down to the byte. What changed is the argument's type, and the enum type stays in the source and never reaches the object file. Your C labels the cases with enum names while the assembly works with their ordinals 0 through 7, and the li r3, N in each arm is the value that case returns.
Your task
Write func_80255b54 to reproduce the assembly above. The Tile enum is provided in context. Eight dense enum values land in table form — read the li r3, N in each arm to recover what each tile costs.