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7.12. strings are mutable

You can change the letters in an string one at a time using the [] operator on the left side of an assignment.

The active code below changes the first letter in greeting to be 'J'.

#include <iostream> using namespace std; int main() { string greeting = "Hello, world!"; greeting[0] = 'J'; cout << greeting << endl; }

This produces the output Jello, world!.

Q-2: What is printed by the following statements?

string fav_food = "ice cream";
fav_food[3] = "d";
cout << fav_food << endl;

• icd cream
• Remember that indexing begins at 0, not 1.
• icedcream
• Index 3 was a space and now it is "d".
• ice cream
• The character at index 3 should be changed to "d".
• iced
• The character at index 3 should be changed to "d", and the rest stays the same.

Q-3: How can we fix the message to be “You’re a wizard Harry”?

string message = "You're a lizard Harry";

• message[9] = "w";
• Since "l" is at index 9, replacing it with "w" fixes the message.
• message[10] = "w";
• Remember indexing starts at 0.
• "w" = message[9];
• In order to change a letter in a string, the ``[]`` operator must be on the left of the assignment.
• message[8] = "w";
• Remember indexing starts at 0.

Put together the code below to create a function mixer that takes in two strings and replaces every even index of the first string by the corresponding index of the second. It returns the modified first string. Example: string_a = "food" and string_b = "summer" . mixer(string_a ,string_b ) makes string_a become “somd”.

Assume second string is greater than first.

        string greeter(string s1,string s2) {
---
void mixer(string s1,string s2) { #paired
---
   int size = s1.length();
---
   int size = s2.length(); #paired
---
   index i = 0;
   while (i < size) {
---
   index i = size - 1; #distractor
   while (i < size) {
---
     if( (i % 2) == 0){
       s1[i] = s2[i];
     }
---
     if( (i % 2) == 1){ #paired
       s1[i] = s2[i];
     }
---
     i++;
---
   }
---
   return s1;
---
   return s2; #paired
---
}
        

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