Two answers, one call

A function hands back one value in x0, but a division has two answers. divmod(a, b, &q, &r) gets both out: the caller (the code making the call) passes the addresses of two of its own variables, and divmod writes the quotient and the remainder into them. It returns 0 in w0 when it worked, and -1 when b is 0, in which case it writes nothing.

The starter's main reads pairs of numbers into its frame until the input runs out, and prints each result from the frame slots q_s and r_s. Write divmod, and fill in the call in main: a in x0, b in x1, and the addresses of the two slots in x2 and x3 (add x2, fp, q_s puts an address in a register).

Use sdivSigned divide; divide-by-zero writes 0., which rounds toward zero the way C does, so -17 / 5 is -3 remainder -2. The remainder is a - q * b, one msubRd = Ra - Rn * Rm.. Check b first: dividing by zero does not stop an ARMv8 program, sdivSigned divide; divide-by-zero writes 0. quietly answers 0.

main exits with the number of pairs that had no answer, so a script running the program can tell something went wrong.

The starter's input prints:

17 / 5 = 3 remainder 2
100 / 7 = 14 remainder 2

What is checked

• the two lines above
• the program exits with status 0
• the same checks on other inputs the checker keeps hidden, including negative numbers, a divisor of 0 (the exit status must count them), and no input at all
• sdivSigned divide; divide-by-zero writes 0. does the division and strStore to memory. Same addressing modes as LDR. writes the answers