The add that is a call
Two floats arrive the way any two 32-bit values arrive: in r0 and r1. The difference is what they contain. Each register holds the raw IEEE-754 bit pattern — sign, exponent, mantissa, packed into a word — and no instruction in the machine can add them.
So the compiler calls someone who can:
0 push {lr}
2 bl __addsf3-4
6 pop {r1}
8 bx r1
The whole function is the call. Arguments are already in the right registers, so there is nothing to set up; __addsf3 reads r0 and r1, returns the sum in r0, and that is also where the return value belongs. The push {lr} and the epilogue exist purely because bl overwrites the link register.
Look at that epilogue closely. With -mthumb-interwork a return is pop then bx, and gcc pops lr into the lowest register that is not carrying a return value. A one-word return leaves r0 occupied, so lr goes to r1. A void function would pop into r0. Before reading a single line of the body you already know how much the function gives back.
Change one character of the source and only the symbol changes:
0 push {lr}
2 bl __subsf3-4
6 pop {r1}
8 bx r1
Same registers, same frame, same size. The operator lives entirely in the helper's name, so translating soft float back to C starts as a lookup: read the symbol, write the operator.
Your target makes two calls, and something has to survive the first one. Work out where it goes and why it has to go there.
Your task
Write func_082841ec to reproduce the target assembly.