Palindrome function assignment solution implemented in C++

The purpose to write a C++ assignment is to gain skill in reading and understanding code that uses iterators. The C++ STL (standard template library) requires iterators in many places, so C++ programmers need to be confident in using iterations.

Complete the provided C++ program by debugging the function isPalindrome.

The function should determine if a vector of ints is a palindrome. For example, {1, 2, 3, 2, 1} and { 1, 2, 3, 3, 2, 1 } are palindromes but { 1, 2, 3, 4, 1 } and { 1, 2, 3, 4, 2, 1 } are not palindromes.

This program uses iterators. The function works in all cases except when an even-length palindrome is passed in.

The output should be:

• False
• False
• True
• true

The given code doesn't work right for the last case and produces this output:

• false
• false
• true
• false

Your job is to figure how the function uses iterators and fix the bug.

#include #include #include #include using namespace std; bool isPalindrome(const vector & v); int main() { vector v1False{ 1, 2, 3, 4, 1 }; cout << boolalpha << isPalindrome(v1False) << endl; vector v2False{ 1, 2, 3, 4, 2, 1 }; cout << boolalpha << isPalindrome(v2False) << endl; vector v1True{1, 2, 3, 2, 1}; cout << boolalpha << isPalindrome(v1True) << endl; vector v2True{ 1, 2, 3, 3, 2, 1 }; cout << boolalpha << isPalindrome(v2True) << endl; system("pause"); return 0; } bool isPalindrome(const vector & v) { if ( v.size() == 0 ) { return true; } auto front = v.begin(); // first element of v. auto rear = --v.end(); // last element of v. // algorithm: // iterators front and rear move towards center. // if there is a mismatch, the loop ends // if palindrome, front and rear both get to middle. // as long as front and rear elements have the same value // and front hasn't gone past rear, increment both. for (; *front == *rear && front < rear; ++front, --rear) { // no op } // check positions of front and rear. // if there is a mismatch, it will happen // before front passes rear. return front >= rear; }