Why do I have this infinite loop - c++

Question

hey guys kind of new to the programming world and I know you should read in all input as a string but this is just a simple program and I'm having a brain fart I think but here is my question......why when I press q to quit am i getting an infinite loop and how would I condense the while loops because that looks gross
here is what I have so far
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int grade;
char quit = 'a';
cout << "Input your grade (0-100): ";
cout << endl;
cin >> grade;
while (grade < 0) {
cout << "If you have a negetive grade....drop out! otherwise enter another grade" << endl;
cin >> grade;
}
while (quit != 'q') {
while (grade < 0) {
cout << "If you have a negetive grade....drop out! otherwise enter another grade" << endl;
cin >> grade;
}
if (grade == 100) {
cout << "You got a perfect grade!" << endl;
cout << "Letter grade: A" << endl;
}
else if (grade >= 90 && grade <= 100) {
cout << "Letter grade: A" << endl << endl;
}
else if (grade >= 80 && grade <= 89) {
cout << "Letter grade: B" << endl << endl;
}
else if (grade >= 70 && grade <= 79) {
cout << "Letter grade: C" << endl << endl;
}
else if (grade >= 60 && grade <= 69) {
cout << "Letter grade: D" << endl << endl;
}
else if (grade < 60) {
cout << "Letter grade: F" << endl << endl;
}
else {
cout << "Invalid grade!" << endl;
}
cout << " would you like to enter another grade? or press q to quit" << endl;
cin >> grade;
}
system("pause");
return 0;`enter code here`
}

Answer 1

Because of grade var. You declared grade as int.
If you type correct int, it works well, but If you type another char ex:) q or f, function cin cannot recognize q or f as int type.
If char input, cin pass own process.
You have to change grade type into char to recognize char and int inputs both.
If you want to use only one input flow, this implementation code will help you.
int _tmain(int argc, _TCHAR* argv[])
{
char c_input[32] = {0};
cin>>c_input;
while(atoi(c_input) < 0)
{
cout<<"If you have a negative grade....drop out! otherwise enter another grade" << endl;
cin>>c_input;
}
while(c_input[0] != 'q')
{
while(atoi(c_input) < 0)
{
cout<<"If you have a negative grade....drop out! otherwise enter another grade" << endl;
cin>>c_input;
}
cout<<c_input;
cout<<"Would you like to enter another grade? or press q to quit" << endl;
cin>>c_input;
}
return 0;
}

Answer 2

Minimal, verifiable example:
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int grade;
char quit = 'a';
cout << "Input your grade (0-100): ";
cout << endl;
cin >> grade;
while (quit != 'q') {
cout << " would you like to enter another grade? or press q to quit" << endl;
cin >> grade;
}
system("pause");
return 0;`enter code here`
}
See the problem with quit?
edit
What I have done is remove (most of) the lines that have nothing to do with quit or the loop.
At this point you should notice that the loop never changes quit.
If you are having trouble with a program, one of the best ways to figure out what is wrong is to get rid of everything that doesn't have anything to do with the error. In time, you'll be able to do this using only your mind. Duuude!
while i'm at it
The correct way to handle user input is to get it as a string, then convert it to what you want.
For example:
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <string>
template <typename T>
T string_to( const std::string& s )
{
T value;
std::istringstream ss( s );
ss >> value >> std::ws;
if (!ss.eof()) throw std::invalid_argument("T string_to()");
return value;
}
int main()
{
std::cout << "Enter a number or 'q': ";
std::string s;
getline( std::cin, s );
if (s == 'q')
{
std::cout << "Good job! You entered 'q'.\n";
}
else
{
try
{
double x = string_to <double> ( s );
std::cout << "Good job! You entered '" << x << "'.\n";
}
catch (const std::exception& e)
{
std::cout << "Foo, you didn't obey instructions and made me " << e.what() << ".\n";
}
}
}