- Which condition checks the result without potentially dereferencing
values.end()? Answer: if (it != values.end() && *it == target) report_match();
Whenever you work with iterator-based searches in C++, your first instinct should be to ask: "could this iterator be past-the-end?" Functions like std::lower_bound always return a valid iterator, but that iterator might point to values.end() — a sentinel position that marks one past the last element. Dereferencing values.end() is undefined behavior, so you must guard against it before you dereference.
The correct approach, C, checks it != values.end() first, then dereferences with *it == target. Because C++ uses short-circuit evaluation with &&, if the first condition is false, the second is never evaluated — so you never dereference an invalid iterator. This is the canonical safe pattern.
Option A uses it <= values.end(), which is meaningless as a guard — every valid iterator satisfies this, including values.end() itself. It doesn't prevent dereferencing the past-the-end iterator at all.
Option B reverses the order of the checks: it dereferences *it before verifying it != values.end(). Even though the end-check is present, it comes too late. If it is already values.end(), you've triggered undefined behavior on the first subexpression.
Option D swaps values.end() for values.begin(), which checks whether the iterator is at the start of the range — completely unrelated to whether it's valid for dereferencing. The iterator could still equal values.end() and this check would pass incorrectly.
The key takeaway: **always put the bounds check before the dereference in a && condition.** Short-circuit evaluation is your safety net — but only if you order your conditions correctly.
- Which loop visits every valid index exactly once without an out-of-bounds access?
Answer: for (std::size_t i = values.size(); i-- > 0; ) use(values[i]);
Iterating a vector in reverse with an unsigned index is a classic C++ trap. The core problem: std::size_t is an unsigned type, meaning it can never be negative — it wraps around to a huge positive number instead. Any loop condition that relies on the index going below zero will produce undefined or infinite behavior.
Option D is the correct approach. The loop initializes i to values.size() (one past the last valid index), then uses the post-decrement expression i-- > 0 as the condition. Here's the key: the post-decrement evaluates the original value of i for the comparison, then decrements it. So on the first iteration, i is compared as size() > 0, and if true, i becomes size() - 1 — a valid index — before the body executes. When i is 1, the condition reads 1 > 0 (true), decrements to 0, and the body uses values[0]. When i is 0, the condition reads 0 > 0 (false), and the loop exits cleanly. If the vector is empty, size() is 0, so 0 > 0 is immediately false — no body execution, no crash.
Option A fails catastrophically on an empty vector: values.size() - 1 underflows to an enormous number, and i >= 0 is always true for unsigned types, creating an infinite loop with out-of-bounds access. Option B accesses values[i] before decrementing, so it reads values[size()] — one past the end — on the first iteration. Option C skips index 0 entirely because i > 0 exits before processing the first element.
Your takeaway: whenever you reverse-iterate with std::size_t, avoid conditions like i >= 0. The post-decrement-in-condition pattern (i-- > 0) is the idiomatic, safe solution.
- Which guard correctly excludes both negative indices and the past-the-end index without unsafe signed-to-unsigned reasoning?
Answer: index >= 0 && std::cmp_less(index, values.size())
When guarding array access with a signed index, you need to solve two problems simultaneously: reject negative values and reject indices that equal or exceed the container's size. The tricky part is that std::vector::size() returns an unsigned type (size_type), so naively mixing signed and unsigned comparisons can produce silent bugs.
The <utility> header in C++20 provides std::cmp_less (and related functions) that perform mathematically correct comparisons between signed and unsigned integers without any implicit conversion. D — index >= 0 && std::cmp_less(index, values.size()) — is the correct guard because it first explicitly filters out negatives with index >= 0, then uses std::cmp_less to safely compare the (now known-positive) signed index against the unsigned values.size(). Both conditions must hold, so past-the-end access is also blocked.
A uses <= instead of <, so it allows index == values.size(), which is a past-the-end access — a classic off-by-one error.
B casts a potentially negative index directly to size_type. A negative value wraps around to a huge unsigned number, which almost certainly passes the <= check, allowing wildly out-of-bounds access — exactly the "unsafe signed-to-unsigned reasoning" the question warns about.
C has the logic backwards: the || means the guard is true (access is allowed) when index < 0, since the second operand isn't even checked. You'd be granting access precisely when you should be blocking it.
Your takeaway: whenever you compare a signed index to an unsigned size, reach for std::cmp_less and pair it with an explicit non-negativity check using &&.
- Which statement most accurately describes this code and the required correction?
Answer: It is invalid because the two iterator arguments come from different temporary vectors; storing one returned vector before searching fixes both range and lifetime issues.
Whenever you see iterators passed to an algorithm, your first instinct should be to ask: do these iterators belong to the same container? Iterators are only meaningful relative to the specific object that produced them — mixing iterators from different containers is undefined behavior, regardless of whether those containers hold identical data.
Here, make_values() is called twice, producing two separate temporary vectors. The .begin() comes from the first temporary and .end() comes from the second. These iterators are not part of the same range. Passing mismatched iterators to std::find is undefined behavior — the algorithm increments from one vector's beginning toward another vector's end, a pairing that has no defined meaning in C++. The fix is straightforward: store the returned vector in a named variable first, then call .begin() and .end() on that single object. This ensures both iterators belong to the same container and that the container outlives the search. That's exactly what A describes, making it correct.
B is wrong because it assumes equal element sequences make iterators interchangeable — they don't. Iterators encode identity (which object, which position), not just value. C incorrectly suggests the code is valid during the search itself; the undefined behavior occurs at the point of the std::find call, not later at dereference. D invents a scenario where the bug only surfaces when 7 is absent — but the iterator mismatch causes undefined behavior unconditionally, regardless of whether the value is found.
As a study tip: any time you see a temporary object's member function chained into an algorithm, ask whether the temporary lives long enough and whether all iterators share the same source.
- Which loop is defined and prints
654? Answer: for (int* p = data + 3; p != data; ) std::cout << *--p;
When working with pointer arithmetic and reverse iteration, the key is tracking exactly where your pointer starts and what it dereferences at each step before moving.
An array int data[] = {4, 5, 6} occupies indices 0–2, so data + 3 points one past the last element — a valid sentinel address you can compare against, but never dereference directly. To print in reverse, you need to step from index 2 down to 0.
Choice A is correct. Starting at data + 3, the expression *--p first decrements p (moving it to data + 2), then dereferences it. So the first iteration reads data[2] = 6, then 5, then 4, printing 654. The loop stops when p == data. Clean, correct, undefined-behavior-free.
Choice B uses *p--, which is post-decrement: it dereferences p before decrementing. On the first iteration, p is still data + 3 — a one-past-the-end address — so dereferencing it is undefined behavior, and the output would be wrong even if it didn't crash.
Choice C also starts at data + 3 but uses --p in the update clause after dereferencing *p in the body. The very first iteration dereferences data + 3, which is again out-of-bounds — undefined behavior before any valid output occurs.
Choice D starts correctly at data + 2 (the last valid element) and iterates down properly, printing 654 — but it uses a >= comparison with a pointer, which is undefined behavior in C++ when the pointer goes below the array's base address.
When you see pointer loops, always ask two things: where does the pointer start, and does it dereference before or after the move? Pre-decrement (--p) inside the dereference is the idiomatic safe pattern for reverse iteration starting one past the end.
- Which algorithm call produces the required result without violating an overlapping-range precondition?
Answer: std::copy_backward(values.begin(), values.end() - 1, values.end());
Whenever you see a question involving copying within an overlapping range, your first instinct should be to ask: which direction does the data need to move, and which algorithm respects that direction safely?
Here, you need to shift elements rightward — copying {1,2,3,4} from positions 0–3 into positions 1–4. The danger with overlapping copies is that the algorithm might overwrite source elements before it reads them. For a rightward shift, you must copy from back to front so that each source element is read before the destination overwrites it. std::copy_backward does exactly this: it copies the range [begin, end-1) — that is, elements {1,2,3,4} — into the destination ending at values.end(), processing from right to left. The result is {1,1,2,3,4}, which is precisely what's required. D is correct.
A uses std::copy in the forward direction on an overlapping rightward shift. std::copy processes left-to-right, so it reads values[0]=1, writes it to values[1], then reads values[1] — which is now 1 instead of 2 — corrupting the output. This is undefined behavior for overlapping ranges where the destination is ahead of the source.
B shifts elements leftward (destination starts at begin), producing {2,3,4,5,5} — the opposite of what's needed.
C uses std::reverse_copy, which copies elements in reversed order into the destination, producing {1,4,3,2,1} — not a shift at all.
Study tip: Remember the rule — rightward shift = copy_backward; leftward shift = copy. Matching copy direction to movement direction keeps you safe from overlapping-range bugs.
- Which revision is guaranteed to avoid using an iterator or reference invalidated by reallocation?
Answer: std::size_t pos = 0; values.push_back(9); std::cout << values[pos];
Whenever you see a question about iterator and reference invalidation, the core rule to remember is this: **any reallocation invalidates all iterators, pointers, and references into a std::vector**. Reallocation happens automatically when you add an element beyond the vector's current capacity. The question tells you size() == capacity(), so the very next push_back or insert will trigger reallocation.
Option C is the safe choice because pos is just a plain std::size_t integer — it's an index, not an iterator or reference into the vector's internal storage. After push_back(9) reallocates the buffer, values[pos] freshly computes the address of element 0 using the new storage. No stale pointer is ever used.
Option A tries to dodge this by calling reserve first, but reserve(capacity() + 1) may itself reallocate — and it does so before saving anything useful. The iterator it is captured before reserve, so it is immediately invalidated by that call, making *it undefined behavior.
Option B stores a int& reference to values.front() before push_back. Since capacity is full, push_back reallocates, and first now dangles — reading it is undefined behavior.
Option D is the same trap as B, just using an iterator instead of a reference. Capturing it before insert (which also triggers reallocation here) leaves it invalidated by the time *it is evaluated.
Study tip: When working with vectors, treat all iterators and references as "expired" the moment any operation could cause reallocation. Indices into a vector are always safe to reuse after reallocation — prefer them when you need stable access across mutations.
- Which change correctly addresses the invalid-iterator defect while retaining the same traversal behavior?
Answer: Remove last and test it != values.end() each iteration, while continuing to assign the iterator returned by erase.
Whenever you see a question about iterator invalidation in C++, the key concept to recall is this: any mutation of a std::vector — including erase — invalidates all iterators at or after the erased element, including end(). That cached last iterator becomes a dangling, undefined reference the moment you erase anything.
The fix in B is correct precisely because it eliminates the stale cache. By calling values.end() fresh on every loop iteration, you always compare against a valid, up-to-date end iterator. The erase call still returns a valid iterator to the next element, so the traversal logic remains intact — nothing about the behavior changes, only the validity guarantee.
A is a tempting but false claim. The C++ standard is explicit: erasing an element invalidates all iterators *from the point of erasure through end()*, inclusive. The end iterator is not magically exempt.
C introduces a second misconception on top of the first. Relational operators (<) do not "tolerate" invalid iterators — using an invalid iterator in any expression, comparison or otherwise, is undefined behavior. Switching from != to < provides no protection whatsoever.
D is a classic pre-C++11 idiom for std::list, where erase does not invalidate other iterators. For std::vector, erase(it++) still post-increments into an invalidated position before the erase occurs, which is undefined behavior.
As a study tip: **always re-evaluate end() inside loops that mutate a std::vector**, and remember that erase's return value is the correct way to advance — not manual incrementing.
- Which return expression has the required result and never calls
front() on an empty vector? Answer: return !values.empty() && values.front() == target;
This question tests your understanding of short-circuit evaluation in C++ logical expressions. When you see a question involving potentially unsafe operations (like accessing an empty container), always ask yourself: "Is there a guard condition, and does it run before the dangerous operation?"
In C++, the && operator evaluates left to right and stops immediately if the left side is false. Similarly, || stops if the left side is true. This short-circuiting behavior is the key to writing safe guard expressions.
Option A is correct because !values.empty() runs first. If the vector is empty, the expression short-circuits to false and values.front() is never called — avoiding undefined behavior. If the vector is nonempty, both conditions are evaluated, returning true only when the first element matches target.
Option B reverses the order — values.front() is called before the empty check. On an empty vector, this triggers undefined behavior regardless of what comes after.
Option C uses ||, which returns true when the vector is empty (since values.empty() is true). This is logically backwards — you'd be returning true precisely when you should return false.
Option D uses || with !values.empty() first. If the vector is nonempty, it short-circuits to true immediately without checking whether the first element equals target — wrong logic entirely.
Study tip: Whenever you need a "guard before access" pattern, reach for && with the safety check on the left. Think of it as a bouncer — the check must come first, or the dangerous code still runs.
- Which loop performs the removal without using an invalidated iterator?
Answer: for (auto it = values.begin(); it != values.end(); ) { if (*it % 2 == 0) it = values.erase(it); else ++it; }
When you erase an element from a std::vector, the iterator pointing to that element is immediately invalidated — using it afterward is undefined behavior. The key to this question is understanding what erase() actually returns: an iterator to the element that now occupies the erased position (or end() if the last element was erased). Safe removal loops must always reassign the iterator using that return value.
Option B is correct because it does exactly this: it = values.erase(it) captures the valid iterator returned by erase, so the loop continues safely from the next element. When the current element is odd, ++it advances normally. Order is preserved throughout.
Option A is the classic trap — it calls values.erase(it) but discards the return value, then implicitly uses it in the next loop check (it != values.end()). Since it is now invalidated, this is undefined behavior, even though it might appear to "work" on some compilers.
Option C tries a clever trick with values.erase(it++), post-incrementing before the erase. While this pattern is safe for node-based containers like std::list (where iterators aren't invalidated by erasure), **it does not work for std::vector** — erasing an element invalidates all iterators at or after that position, including the already-incremented one.
Option D has the increment ++it in the loop header, which means after an erase, it is set to the next valid element by the return value, but then the loop header increments it again, skipping an element entirely.
Study tip: For std::vector erasure loops, always use it = container.erase(it) — treat the return value as your only safe path forward.
