Why is exit(0); giving me a std:string... error? - c++

Question

I'm new to C++. I decided to not watch the next tutorial and put my skills to use, by making a funny Mind Reader application. I'm pleased with myself, however, even though I've ironed out most bugs, I still have one concerning the exit function. I read the C++ documentation for it, and I'm not sure what I did wrong. I did exit(0);. I have a very weird error, which is:
no match for call to '(std::string {aka std::basic_string<char>}) (int)
I have searched online, however I am still unaware of what the problem is. My error is on line 59 (marked in the code):
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main()
{
//declaring variables to be used later
string name;
string country;
int age;
//header goes below
cout << "#######################################";
" ############ MIND READER ############"
"#######################################\n\n";
//asks if the user would like to continue and in not, terminates
cout << "Would like you to have your mind read? Enter y for yes and n for no." << endl;
cout << "If you do not choose to proceed, this program will terminate." << endl;
string exitOrNot;
//receives user's input
cin >> exitOrNot;
//deals with input if it is 'y'
if (exitOrNot == "y"){
cout << "Okay, first you will need to sync your mind with this program. You will have to answer the following questions to synchronise.\n\n";
//asks questions
cout << "Firstly, please enter your full name, with correct capitalisation:\n\n";
cin >> name;
cout << "Now please enter the country you are in at the moment:\n\n";
cin >> country;
cout << "This will be the final question; please provide your age:\n\n";
cin >> age;
//asks the user to start the sync
cout << "There is enough information to start synchronisation. Enter p to start the sync...\n\n";
string proceed;
cin >> proceed;
//checks to see if to proceed and does so
if (proceed == "p"){
//provides results of mind read
cout << "Sync complete." << endl;
cout << "Your mind has been synced and read.\n\n";
cout << "However, due to too much interference, only limited data was aquired from your mind." << endl;
cout << "Here is what was read from your mind:\n\n";
//puts variables in sentence
cout << "Your name is " << name << " and you are " << age << " years old. You are based in " << country << "." << endl << "\n\n";
cout << "Thanks for using Mind Reader, have a nice day. Enter e to exit." << endl;
//terminates the program the program
string exit;
cin >> exit;
if (exit == "e"){
exit(0); // <------------- LINE 59
}
}
}
//terminates the program if the input is 'n'
if (exitOrNot == "n"){
exit(0);
}
return 0;
}
Thanks

Answer 1

The local variable exit shadows other identifiers from outer scopes with the same name.
To illustrate with a smaller example:
int main()
{
int i;
{
int i;
i = 0; // assign to the "nearest" i
// the other i cannot be reached from this scope
}
}
Since the only exit visible is an object of type std::string, the compiler sees exit(0) as a call to operator()(int) and throws a hissy fit when it doesn't find one among std::string members.
You can either qualify the name (std::exit(0);) or rename the variable. And since all of your code is in main you can simply say return 0; instead.

Answer 2

Try using return 0; or return EXIT_SUCCESS;. It's the exact same thing. Also, you can only input one word into a cin. Instead, use getline(cin, string name); If it still doesn't work, add a cin.ignore(); before your getline(cin, string name);, like this:
//stuff
string country;
cout << "Now please enter the country you are in at the moment:\n\n";
cin.ignore();
getline(cin, country);
//stuff
return 0;

Answer 3

The problem is arrising because you declared a standard keyword as the name of a local variable.
Now as the local variable is of type sting it is not able to take it as its value.