Order of While Loops - c++

Question

So, I am fairly new to C++, and I am basically just trying to figure out how I would make things more user-friendly.
I made a fairly basic "calculator" type of program just to mess around!
It uses a lot of while loops for verification to make sure that if they make a mistake, they can re-input the data they want.
I am wondering, once it "outputs" their number, and the program ends, how would I be able to basically just output and then give them the option to restart (from the beginning).
Not sure if I explained that very clearly, but I am fairly new so I have no clue how I would do this seeing on how it is being read top to bottom.
Code:
int a;
int b;
string c;
cout << "1. Addition" << endl << "2. Subtraction" << endl << "3. Multiplication" << endl << endl;
cin >> c;
cout << endl;
while (c != "1" && c != "2" && c != "3"){
cout << "1. Addition" << endl << "2. Subtraction" << endl << "3. Multiplication" << endl << endl;
cin >> c;
cout << endl;
}
while (c == "1"){
cout << "Number 1: ";
cin >> a;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 1: ";
cin >> a;
}
cout << "Number 2: ";
cin >> b;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 2: ";
cin >> b;
}
cout << endl << "Output: " << a + b;
c = "kevin123";
cin.get();
}
while (c == "2"){
cout << "Number 1: ";
cin >> a;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 1: ";
cin >> a;
}
cout << "Number 2: ";
cin >> b;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 2: ";
cin >> b;
}
cout << endl << "Output: " << a - b;
c = "kevin123";
cin.get();
}
while (c == "3"){
cout << "Number 1: ";
cin >> a;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 1: ";
cin >> a;
}
cout << "Number 2: ";
cin >> b;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 2: ";
cin >> b;
}
cout << endl << "Output: " << a * b;
c = "kevin123";
cin.get();
}
cin.get();
return 0;

Answer 1

I would say that your code need a bit of tidying up and refactoring. It repeats a lot of code, so you you should probably rearrange it to avoid that repetition, so for example:
cout << "Number 1: ";
cin >> a;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 1: ";
cin >> a;
}
cout << "Number 2: ";
cin >> b;
while (!cin) {
cout << endl;
cin.clear();
cin.ignore(256, '\n');
cout << "Number 2: ";
cin >> b;
}
occurs identical for alternatives 1, 2 and 3.
You could make that into a common section, and only do the final step of the math to produce the result in different pieces of code. This should be done with if (c == "1") ... etc, since you don't actually want to loop in this case (or use a switch, since you are never going to have both c=="1" and c=="2" at the same time, so a switch on c[0] would be fine...). This would get rid of the c = "kevin123"; which is just an ugly hack to fix the problem that you are looping, when you don't really want to repeat things.
All of your validation loops could also be written as
do { cout << "prompt"; cin >> input; } while(...)
instead of repeating the prompt and reading of input.
Once you have made those modifications, you should wrap the entire main body of your program in another while or do { ... } while(...) loop to repeat the main body.
You could also make use of functions to perform reduce repetitions in the code.
It is a very good practice to learn from early on in programming to "do not repeat yourself" - do not write code that is identical or nearly identical - break the code out into functions, move it around so it's not repeating. This pays off in the long term, since when you need to change something, there is only one or a few places to change, rather than lots!

Answer 2

What your are looking for is a conditional outer most loop. You have is while loop for all possible values. This code is something you can work with.
int a, b;
char c;
cout << "Welcome To Calculator" << endl;
cout << "Please Enter\n1.Addition\n2.Substraction\n3.Division\n4.Multiplication" << endl;
cout << "Enter 'q' for exit" << endl;
while(cin >> c) {
if (isdigit(c))
{
if (c == '1')
cout << "Looking for Addition" << endl;
else if (c == '2')
cout << "Looking for Substraction" << endl;
else if (c == '3')
cout << "Looking for Division" << endl;
else if (c == '4')
cout << "Looking for MUltiplication" << endl;
// Requesting new calculation.
cout << "Please Enter\n1.Addition\n2.Substraction\n3.Division\n4.Multiplication" << endl;
cout << "Enter 'q' for exit" << endl;
}
else
break;
}
cout << "Thankyou for using the calculator" << endl;
return 0;
#user3646954:
The problem with your code is not only about the repetition of code but also the algorithm.
while (c != "1" && c != "2" && c != "3") checks for right the option (1 | 2 | 3). After which you loop around the options while( c == '1 | 2 | 3'). The loop doesn't makes sense since the loop will always exit after the first calculation (c = "kevin123"; cin.get()). You are never restarting your calculator.
The code snippet above removes the repetition of code for all options. The operator is requested at the start while(cin >> c). After completing the calculation ìf .. else if .. else if .. we go back to the top. (check the code).