Exercises — break, continue, pass — when and why

break. "Stop the whole job right now" = leave the loop entirely. continue would only skip the current item and keep scanning; pass would do nothing at all. Look at figure s01: break is the arrow that leaves the cycle.

continue skips the print only for i == 2. Every other value prints. Output:

0
1
3
4

pass does nothing each cycle, so the loop simply runs to completion. The loop variable c keeps its last value, "h" then "i", ending at "i". Output:

i

total = 0
for n in range(1, 11):
    if n % 4 == 0:
        continue          # skip 4 and 8
total += n            # (this line is inside the loop)

Corrected, properly indented:

total = 0
for n in range(1, 11):
    if n % 4 == 0:
        continue
    total += n
print(total)

Skipped values: $4,8$. Kept: $1+2+3+5+6+7+9+10=43$. Total = 43.

nums = [3, 8, 2, 15, 6]
for x in nums:
    if x > 10:
        print(x)
        break
else:
    print("none")

15 is the first value over 10, so it prints and break fires. Because a break happened, the else is skipped. Output:

15

Walk the cycle for each i:

• i = 0: even → continue → nothing printed.
• i = 1: odd, not > 3 → prints 1.
• i = 2: even → continue.
• i = 3: odd, not > 3 (3 is not > 3) → prints 3.
• i = 4: even → continue (the break check is below the continue, so it's never reached for even numbers).
• i = 5: odd, and 5 > 3 → break. Loop ends.

Output:

1
3

Note the subtlety: i = 4 would have triggered break if the order were reversed — but continue fires first for even numbers, so 4 never reaches the break test.

break exits only the nearest enclosing loop — the inner b loop. The outer a loop keeps going.

• a = 0: b = 0 prints 0 0; b = 1 → break (inner loop ends).
• a = 1: b = 0 prints 1 0; b = 1 → break.
• a = 2: b = 0 prints 2 0; b = 1 → break.

Output:

0 0
1 0
2 0

n = 21
i = 2
while i * i <= n:
    if n % i == 0:
        print("not prime")
        break
    i += 1
else:
    print("prime")

Trace: i = 2, 2*2 = 4 <= 21, 21 % 2 = 1 → not a divisor, i becomes 3. i = 3, 9 <= 21, 21 % 3 = 0 → divisor found! Prints "not prime", break. The else is skipped. Output:

not prime

(See Prime Number Checking for the full pattern and why i*i <= n is enough — no divisor above $\sqrt{n}$ can exist without a matching one below it.)

def handle(event):
    if event == "quit":
        print("bye")
    elif event == "save":
        pass            # TODO: implement saving
    else:
        print("unknown")

• handle("save") → the pass branch runs, does nothing, function returns silently. Prints nothing.
• handle("open") → "open" matches neither → else → prints "unknown".

pass here keeps the elif block syntactically legal while you defer its logic — see Functions and def and Conditionals (if-elif-else).

No value in [1, 3, 5, 7] is even, so the if never triggers, break never fires, and found_even stays False. The continue on the last line simply sends us to the next iteration each time — it does not suppress the else. The loop reaches its natural end with no break, so the else runs. Output:

all odd
False

Key insight: only break suppresses else. continue is irrelevant to the else decision — see for...else and while...else.

data = [4, 9, 16, 25, 30, 36]
for v in data:
    if v % 2 != 0:
        continue                  # skip odd numbers early
    root = int(v ** 0.5)
    if root * root == v:          # v is a perfect square
        print(v)
        break
else:
    print("none")

Trace:

• 4: even, int(4**0.5)=2, 2*2=4 ✓ → prints 4, break.

The very first element already qualifies. Output:

4

The continue keeps the even-filter at the top so the perfect-square test reads cleanly without nesting — exactly the design goal from the parent note.

• break = "leave the building." Nearest loop only.
• continue = "skip this room, go to the next." Loop keeps running; else still fires at the end.
• pass = "stand here doing nothing, then carry on." Purely for syntax.