Walkthrough: Sign of a Number

例题: 数的正负号

≈ 8 min · python-conditionals · Open in WolfCode

Three mutually exclusive outcomes → if / elif / else. The simplest possible 'chained conditional' from Think Python §5.5, end-to-end.

The problem

Write sign(n) so that it returns "positive", "negative", or "zero" depending on n. This is a worked example (kind = example): read it, run it, modify it. The autograder will pass without any changes.

The pattern

Click ▶ Run — predict the output first

def sign(n):
    if n > 0:
        return "positive"
    elif n < 0:
        return "negative"
    else:                       # only reachable when n == 0
        return "zero"

for n in [5, -3, 0, 100, -0.0001]:
    print(f"sign({n!r:>9}) = {sign(n)}")

Loading Python runtime (~5 MB, one-time)…

Why not three separate `if`s?

A common beginner version works only because each branch has a return — but it does extra checks and is easier to break with a refactor:

Works but wastes effort

def sign_buggy(n):
    if n > 0:
        return "positive"
    if n < 0:                # always checked, even when n > 0
        return "negative"
    if n == 0:
        return "zero"

print(sign_buggy(5))
print(sign_buggy(-3))

Loading Python runtime (~5 MB, one-time)…

The elif version makes the intent explicit: these are alternatives, not three independent questions.

Trace it by hand

n = 5 → first condition true → returns "positive"
n = -3 → first false, second true → returns "negative"
n = 0 → both conditions false → else fires → returns "zero"

Open the next lesson — Letter Grade — which uses the exact same pattern but with five branches and an order-of-branches trap.