Letter Grade

等级字母

≈ 15 min · python-conditionals · Open in WolfCode

Same elif chain as the example — but with five branches and an order-of-branches trap that bites almost everyone the first time.

Specification

Implement letter_grade(score):

score >= 90 → "A"
score >= 80 → "B"
score >= 70 → "C"
score >= 60 → "D"
score < 60 → "F"

The trap

What does this return for score = 95? Predict first.

def letter_grade(score):
    if score >= 60:
        return "D"
    elif score >= 70:
        return "C"
    elif score >= 80:
        return "B"
    elif score >= 90:
        return "A"
    else:
        return "F"

print(letter_grade(95))
print(letter_grade(72))
print(letter_grade(50))

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

The fix: order high-to-low

Python · runnable

def letter_grade(score):
    if score >= 90:
        return "A"
    elif score >= 80:
        return "B"
    elif score >= 70:
        return "C"
    elif score >= 60:
        return "D"
    else:
        return "F"

# Boundary tests — these matter to the autograder:
for s in [95, 90, 89, 80, 70, 60, 59, 0]:
    print(f"letter_grade({s:>3}) = {letter_grade(s)}")

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

Check yourself

Read Think Python §5.5 again if this surprises you.

With the BROKEN ordering above, letter_grade(95) returns:

Now you try in WolfCode

Open 01-letter-grade.py and write the correct version. The autograder checks all 8 cases including 90, 60, and 59 (boundaries). The Tutor is enabled at L1–L3 — L4 is off by design.