Exercises — Control flow — if - else, switch, while, do-while, for, break, continue, goto

WHAT we do: check each condition top to bottom. WHY: an if/else if/else chain stops at the first true branch and skips the rest.

• n > 0 → 0 > 0 → false.
• n == 0 → 0 == 0 → true → print B, then skip the else.

Answer: B

A while tests before the body. First test: 10 < 3 → false → body runs zero times. Answer: done

A do-while runs the body first, then tests. So it prints 10 once, i becomes 11, test 11 < 3 false → stop. Answer: 10 done

switch jumps to the matching label then falls through everything below until a break — and there are no breaks here. Landing on case 2, execution runs case 2, case 3, and default in a row. Answer: two three end

WHAT: build the 4-part loop (init, condition, update) with the body doing the print. WHY for: we know exactly how many times we count, so a counting loop is the natural fit.

for (int i = 0; i < 5; i++) {
    printf("%d ", i);
}

Condition i < 5 (strict less-than) stops after i = 4. Output: 0 1 2 3 4.

int sum = 0;
for (int i = 1; i <= 100; i++) sum += i;
printf("%d", sum);

WHY i <= 100: we want to include 100, so we use less-than-or-equal. The famous closed form $\frac{n(n+1)}{2}=\frac{100\cdot 101}{2}$ tells us the answer must be 5050 — a good self-check.

WHY do-while: we must ask once before we can judge the answer — "do, then decide whether to repeat" is exactly the do-while shape.

int choice;
do {
    printf("Enter 1-3: ");
    scanf("%d", &choice);
} while (choice < 1 || choice > 3);

The || is logical OR: the loop repeats if choice is below 1 or above 3.

Trace each i:

• i=0: not 1, not 4 → print 0.
• i=1: continue → jump straight to i++ (update), print nothing.
• i=2: print 2.
• i=3: print 3.
• i=4: break → leave the loop entirely.

Answer: 0 2 3

x = 5 is assignment, not comparison. It stores 5 into x, and the whole expression's value is 5. Since 5 is non-zero, the condition is true → prints yes — always, regardless of x's old value. The author almost surely meant x == 5 (comparison). Answer: yes (buggy).

WHAT goes wrong: at i == 2, continue jumps back to the condition test — but i++ sits after the continue, so i is never incremented. i stays 2 forever → infinite loop (prints 0 1 then hangs). WHY the for version was safe: in a for, continue still runs the header's update i++. In while you must run the update yourself before any continue. Fix — increment before the skip:

int i = 0;
while (i < 5) {
    int cur = i;
    i++;                 // update first
    if (cur == 2) continue;
    printf("%d ", cur);
}                        // prints 0 1 3 4

Output after fix: 0 1 3 4.

WHY this order of checks: test "both" first (% 3 == 0 && % 5 == 0), because if you test 3 alone first you'd print Fizz and miss the FizzBuzz case. The % operator (see C Operators) gives the remainder; remainder 0 means "divides evenly."

for (int i = 1; i <= 15; i++) {
    if (i % 3 == 0 && i % 5 == 0) printf("FizzBuzz\n");
    else if (i % 3 == 0)          printf("Fizz\n");
    else if (i % 5 == 0)          printf("Buzz\n");
    else                          printf("%d\n", i);
}

Trace of the interesting cells: 3→Fizz, 5→Buzz, 15→FizzBuzz. The count of Fizz-only outputs in 1..15 is at 3,6,9,12 → 4 of them.

WHY goto here: a plain break only exits the inner loop; we'd re-enter the outer loop and keep scanning. goto jumps clean out of both at once — the one widely-accepted use of goto.

int grid[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int target = 7, r = -1, c = -1;
for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
        if (grid[i][j] == target) { r = i; c = j; goto found; }
found:
printf("(%d,%d)", r, c);

Scanning row 0 (1 2 3), row 1 (4 5 6), row 2 starts with 7 → match at i=2, j=0. Answer: (2,0)

'X' matches no case, and there is no default. In that situation switch simply does nothing and control continues after the block. So only the final printf runs. Answer: |end (If a default existed, it would have caught 'X'.)

The trap: many expect -7 % 2 to be 1. In C (since C99) the result of % takes the sign of the dividend (n), so -7 % 2 == -1, not 1. But we only test == 0, and -1 != 0, so the branch is "odd" — correct here. Answer: odd — but be warned: if you had written if (n % 2 == 1) to detect odd, it would fail for negatives because -7 % 2 is -1, not 1. Safer parity test: if (n % 2 != 0).

The condition is the constant 0 → false. But a do-while runs the body once before testing, so tries becomes 1, then while(0) stops. Answer: 1 This do { ... } while(0) is a classic C idiom for a single-pass block you can break out of (used in macros and cleanup chains) — it guarantees exactly one execution while still allowing early break.

The trap: the ; right after the for(...) makes the loop body empty — the loop just runs its update until the condition fails. It increments i from 0 while i < 5. When does it stop? At the moment i becomes 5, the test 5 < 5 is false, so it stops with i == 5. Answer: 5 The loop "did nothing" five times, but its side effect left i one past the last valid index.

A switch lands on a matching case and does what without break? ::: Falls through into every following case's code until a break or the block end. do { } while(0) runs the body how many times? ::: Exactly once (body runs before the false test). In C99, -7 % 2 equals? ::: -1 (the remainder takes the sign of the dividend). Sum of 1..100? ::: 5050. for (i=0; i<5; i++); leaves i equal to? ::: 5. Why does a lone break fail to exit nested loops? ::: It exits only the innermost loop; the outer loop continues.