Handling Mismatch Results

When using std::ranges::mismatch(), it's important to handle past-the-end iterators properly to avoid dereferencing invalid iterators. std::ranges::mismatch() returns a pair of iterators, indicating where the first mismatch occurs.

If no mismatch is found, both iterators will be past-the-end iterators for their respective ranges. Here's an example demonstrating this:

1#include <algorithm>
2#include <vector>
3#include <iostream>
4
5int main() {
6  std::vector<int> A{1, 2, 3};
7  std::vector<int> B{1, 2, 4};
8
9  auto [itA, itB] = std::ranges::mismatch(A, B);
10
11  if (itA != A.end() && itB != B.end()) {
12    std::cout << "First mismatch - A: " << *itA
13              << ", B: " << *itB; 
14  } else {
15    std::cout << "No mismatch found,"
16      " or one range is longer";
17  }
18}

1First mismatch - A: 3, B: 4

If one of the ranges is longer, the past-the-end iterator indicates that the comparison reached the end of the shorter range without finding a mismatch. You can handle this scenario as follows:

1#include <algorithm>
2#include <iostream>
3#include <vector>
4
5int main() {
6  std::vector<int> A{1, 2, 3};
7  std::vector<int> B{1, 2, 3, 4};
8
9  auto [itA, itB] = std::ranges::mismatch(A, B);
10
11  if (itA == A.end()) {
12    std::cout << "Reached the end of A";
13  }
14
15  if (itB != B.end()) {
16    std::cout << "\nMismatch in B: " << *itB;
17  }
18}

1Reached the end of A
2Mismatch in B: 4

In this example, we check if itA is equal to A.end(), indicating we've reached the end of the first range. Similarly, we check if itB is not equal to B.end() to identify a mismatch in the second range.

Handling past-the-end iterators correctly ensures that your program doesn't attempt to access invalid memory, which can lead to undefined behavior or crashes. Always verify that iterators are within valid ranges before dereferencing them.