C++ Foundations — Quiz Answers

The explanations are the lesson · Quiz · Module home

Q1 — b. The compiler meets #include lines first (preprocessing), then function definitions — main among them — whose statements execute when the program runs, ending at return. (a) and (c) put includes after use; (d) lists comments as a structural part — comments are removed before compilation.

Q2 — c. “Exactly one main” is the contract; a second produces error: redefinition of 'int main()'. Nothing runs — the build refuses. (A linker error, formally: both definitions fight for the entry point.)

Q3 — b. main’s return value goes to the operating system; 0 is the conventional success code. Nothing about memory or line counts.

Q4 — c. 2ndTotal starts with a digit — the one identifier rule beginners actually break. _hidden is legal (though discouraged by convention); MAX_MARKS is a fine constant name.

Q5 — c. A literal with a decimal point is a double. '3.14' would be an invalid char; "3.14" a string; no cast involved.

Q6 — b. Assignment copies the value at that moment. y is an independent box; changing x afterwards never reaches it. (Linking is what references do — a later-course idea.)

Q7 — b. The three benefits in one line: self-documenting name, single update point, compile-time protection against accidental change. (a) is false as a general claim; (c) is backwards; (d) is cosmetic.

Q8 — b. 7 / 2 sees two ints → integer division → 3, printed as 3. (3.5 needs a cast or a decimal-point operand — Lesson 4’s crown jewel.)

Q9 — b. Precedence: arithmetic before comparison → (2+3) < (5+1) → 5 < 6 → true → promoted to 1 → 1 + 1… wait — check that again! true is 1, so 1 + 1 = 2? No — re-read the expression: it’s (2 + 3 < 5) + 1: (2+3 < 5) is 5 < 5 → false → 0; 0 + 1 → 1. The printed answer is 1 — but if you computed 5 < 6 you parsed the outer + 1 as inside the comparison. The expression as written has + 1 outside the parentheses, so the comparison is 5 < 5. This is exactly why the course parenthesises everything: the answer key’s job is to make the parsing unambiguous — and this question is the drill.

Q10 — c. Postfix: b receives the old 5, then a becomes 6. (++a would have given 6 and 6.)

Q11 — c. The cast happens before the division, so the division runs in doubles: 17.0 / 2 = 8.5. (8 would be uncast; 8.0 is cast- after-the-fact; 9 is rounding — nothing rounds here.)

Q12 — b. = stores 50 (truthy) into marks and the condition is always true — the assignment-in-condition bug, Lesson 4’s M4. -Wall warns; the fix is ==.

Scoring: 1 point each. 11–12 → module complete: on to the Lab if not already done, then Unit 04. 8–10 → re-read the flagged sections and retake tomorrow. ≤ 7 → redo exercises + predictions + debugging before retaking. (bands)

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