Question

Could anyone help?

Thank you.

B.6 - 10 pts string* callMe(string& s) { s = "Senorita"; return &s; int main() { string s = "Senor"; string* str = callme(s); cout << "Call me ""+ *str + "\""; return; What is the output or the error of this program? And please explain the data flow in detail (to memory level).

Answer 1

output : -

call me 'Senorita'

program with explanation :-

#include<iostream> // for input output
#include<string> // for string class
using namespace std; //for cin cout

// assigns "Senorita" this value to passing string argument i.e it overides at return the same string
string* callme(string& s){
s = "Senorita";
return &s;
}

// entry point of the program
int main(){
// declare string s with value "Senor"
string s = "Senor";
  
// assign return value of callme function to string str
string* str = callme(s);
  
// concate and prints the output
std::cout << " call me \'" + *str + "\'";
  
return 0;
}

Flow wise explanation : -

1. program execution starts from main() function

2. string s = "Senor"; this expression assigns value "Senor" to suppose xyz location

3. string* str = callme(s); this expression (only rigth part execute now) calls function callme() with pointer to string s.No memory updation

4. s = "Senorita"; this expression overides value "Senorita" to xyz location. It means it changes the value at this location.

5. return &s; this expression returns address of string s i.e xyz

6. string* str = callme(s); (left part execute now) here str points to the location xyz now i.e str is string pointer to xyz location

7. " call me \'" + *str + "\'" this expression concates output with value represeting by string pointer str

8. std::cout << it prints output to the console

6.