The bigger of two numbers
A compare and a jump pick the larger of two registers, and a second jump turns the difference into a distance.
Two numbers sit in registers, and the program leaves the larger of them in d and the distance
between them in e. Both answers come out of the same pair of instructions, a cp that subtracts
and keeps the flags and a jr that reads them.
The two programs before this one ran every instruction they had, top to bottom. This is the first one where some instructions are skipped, and stepping through it is how you watch which ones.
You need to know: the "jp, jr and the conditions" lecture and the "The F register" lecture. What is new here is that a jump chooses between two pieces of code, so the piece that runs first has to jump over the one that follows it.
cp c computes a - c and throws the answer away, so it needs x in a first: the accumulator is
the only register cp compares against, which is why the ld a, b above it is there and why it
comes back a second time before the subtraction.
C is set when the subtraction had to borrow, which is exactly when a was the smaller of the two,
so jr nc under the cp means "if x is greater than or equal to y". That is the unsigned
comparison. The M68K writes bge here and gets the signed one for the same price; the Z80 has no
signed condition at all, and the five instruction S against P/V test from the F register lecture
is what a program writes when its numbers can go below zero. These two cannot, so jr nc is the
whole test.
The jr done is the difference between the two halves. An if with an else has two pieces of
code and only one of them may run, so the first one ends by jumping over the second; leave the jr
out and the program falls through into x_is_bigger and overwrites the answer it just wrote. An
if with no else, like the neg below it, has nothing to jump over.
d comes out at 40, which is 64, and e at 1B, which is 27, so the panel shows de as 401B.
The sub c sets C itself, so no second comparison is needed before the jr nc: an instruction
that computed something has already said whether it borrowed.
Try changing ld b, 37 to ld b, 99. de comes out at 6323, which is 99 and 35, and the two
jumps that were not taken are now the ones that are.