使用指针的C ++反转字符串

Question

I have been given the task to receive a string input from the user and reverse the order of the string and print the result out. My code is this:

#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int main() {
    string input;
    char *head = new char, *tail = new char;
    char temp;
    //Get the string from the user that will be reversed
    cout << "Enter in a string that you want reversed: ";
    getline(cin, input);
    //Create and copy the string into a character array
    char arr[input.length()];
    strcpy(arr, input.c_str());
    //Set the points of head/tail to the front/back of array, respectably
    head = &arr[0]; tail = &arr[input.length()-1];
    //Actual reversal part of the code (Does not work)
    for(int i=0; i<input.length(); i++) {
        temp = *(tail);
        *tail = *head;
        *head = temp;
        tail --; head ++;
    }
    //Print the character array
    for(int i=0; i<input.length(); i++) {
        cout << arr[i];
    }
    //Free up memory
    delete head; delete tail;
    head = NULL; tail = NULL;
    return 0;
}

When I print it, literally nothing has been changed and I can't seem to understand why as I'm brand new to pointers. This is the specific block that I'm having trouble with:

    for(int i=0; i<input.length(); i++) {
        temp = *(tail);
        *tail = *head;
        *head = temp;
        tail --; head ++;
    }

Any input on how to fix this or pointer knowledge in general that'd help is greatly appreciated.

Answer 1

Your approach is good but...

for(int i=0; i<input.length(); i++) {
        temp = *(tail);
        *tail = *head;
        *head = temp;
        tail --; head ++;
    }

Didn't you try working this out on paper? You swap each pair of letters twice , bringing the array back to its original order.

Just change the limit of iteration, to stop when head and tail meet in the middle, and you'll be all right:

for(int i=0; i<input.length()/2; i++) {
  ...
}