Off-by-one check

Questions 1 to 5 of 6

1. 1. This code should print the three names at positions 2, 3 and 4 (Chen, Dara and Eli). Which statement is correct? ``cpp #include <iostream> #include <string> #include <vector> using namespace std; int main() { vector<string> names = {"Ada", "Ben", "Chen", "Dara", "Eli", "Fay", "Gus", "Hana"}; for (int i = 2; i < 4; ++i) { cout << names[i] << "\n"; } return 0; } ``

   The start should be 1, because positions in a vector begin at 1.names[i] should be names.at(i - 1), because the first name is at position 1.The end is written as 4, but positions 2, 3 and 4 form the half-open range [2, 5), so the condition should be i < 5.The loop runs too many times, so it should stop at i < 3.
2. 2. A calendar stores each meeting as a half-open range, so 9 to 10 ends at 10 and 10 to 11 starts at 10. This program checks for a clash. Which statement about it is correct? ``cpp #include <iostream> using namespace std; struct Meeting { int start; int end; }; bool overlaps_closed(Meeting a, Meeting b) { return a.start <= b.end && b.start <= a.end; } int main() { Meeting first = {9, 10}; Meeting second = {10, 11}; cout << "clash: " << overlaps_closed(first, second) << "\n"; return 0; } ``

   It prints clash: 1, because the <= test counts the hour 10 as busy for both meetings; half-open ranges need < in both comparisons.It prints clash: 0, because 10 is not greater than 10 in either comparison.It does not compile, because a struct cannot be passed to a function.It prints clash: 1, but only because the two meetings have different lengths.
3. 3. This code should print each of the 8 names once, in groups of three. Which statement is correct? ``cpp #include <algorithm> #include <iostream> #include <string> #include <vector> using namespace std; int main() { vector<string> names = {"Ada", "Ben", "Chen", "Dara", "Eli", "Fay", "Gus", "Hana"}; int n = static_cast<int>(names.size()); for (int lo = 0; lo < n; lo += 3) { int hi = min(lo + 3, n - 1); for (int i = lo; i < hi; ++i) cout << names[i] << " "; cout << "\n"; } return 0; } ``

   The loop should start at lo = 1, because position 0 is not a name.The inner loop should use i <= hi, so that position hi is printed as well.The outer loop should use lo <= n, so that the group starting at position 6 is reached.The end of each group should be min(lo + 3, n), because the last group must end at the list size. With n - 1, Hana at position 7 is never printed.
4. 4. What does this program print? Each group is one line. ``cpp #include <algorithm> #include <iostream> #include <vector> using namespace std; int main() { vector<int> v = {1, 2, 3, 4, 5, 6, 7}; int n = static_cast<int>(v.size()); for (int lo = 0; lo < n; lo += 3) { int hi = min(lo + 3, n); cout << "[" << lo << ", " << hi << ") " << hi - lo << "\n"; } return 0; } ``

   [0, 3) 3 ; [3, 6) 3 ; [6, 8) 2[0, 3) 3 ; [3, 6) 3 ; [6, 7) 1[0, 3) 3 ; [3, 6) 3[0, 3) 3 ; [3, 6) 3 ; [6, 9) 3
5. 5. What does this program print? Each number is on its own line. ``cpp #include <algorithm> #include <iostream> #include <vector> using namespace std; int main() { vector<int> v = {1, 3, 3, 8, 9}; cout << lower_bound(v.begin(), v.end(), 3) - v.begin() << "\n"; cout << lower_bound(v.begin(), v.end(), 10) - v.begin() << "\n"; return 0; } ``

   1 then 43 then 51 then 52 then 6