Reading a chain, then the constant on the end
Scale a value and offset it and you have most of the arithmetic in a game: a tile index into an address, a frame counter into a table position, a percentage into a bar width. On this compiler that is a shift chain followed by one immediate instruction, and the two halves have to be read separately.
Here is x * 17 - 50:
0 mov r1, r0
2 lsl r0, r1, #4
4 add r0, r1
6 sub r0, #50
8 bx lr
The first three instructions are the multiply you already know — copy the multiplicand aside, shift by 4 for sixteen copies, add the original for seventeen. The fourth is the offset, and it is recognisable because it carries an immediate rather than a register: the chain only ever refers back to the preserved multiplicand.
Longer chains work the same way if you track the running factor line by line:
lsl rD, #n multiplies the factor you have so far by 2^n;add rD, rM where rM is the preserved original adds one to the factor;sub rD, rM against the original subtracts one from it.
Start the count at 1 and apply each line in order. A chain ending in a shift is an even constant — gcc factors it as an odd number times a power of two, builds the odd part first, and shifts at the end.
The trap is deciding where the multiply stops and the offset starts. A trailing lsl belongs to the scale; the immediate add or sub after it is the offset. Get that boundary wrong and you will be looking for a constant that does not exist.
Count your target's chain out to its factor before you go anywhere near the last line.
Your task
Write func_080782e0 to reproduce the target assembly.