Yes/No program using while loop in C++ [closed] - c++

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I'm trying to make a program that asks the user whether they would like to continue to the next calculation. For some reasons whenever I enter y or Y, the program ends. However, if I use only one condition inside the if statement(without '||'sign), the code works fine. I just want to make sure that the user can enter both upper and lower case.
What's wrong with the code? Is there a better way to do this?
int main()
{
char choice;
while(true)
{
cout<<"Would you like to perform other calculation?(Y/N)"<<endl;
cin >> choice;
if(choice == 'Y'|| choice =='y'){
return true;
}else if(choice =='N'||choice =='n'){
return false;
}
}
return 0;
}

Answer 1

The return statement ends a function, in this case, this is the main, so it ends your program, whatever value you're returning.
If you only want to get out of your loop you have two solutions:
Use a boolean:
int main()
{
char choice;
bool run = true; //#stefaanv
while(run)
{
// Make your calculation
cout<<"Would you like to perform other calculation?(Y/N)"<<endl;
cin >> choice;
if(choice =='N'||choice =='n'){
run = false;
}
}
return 0;
}
Use break to exit your loop:
int main()
{
char choice;
while(true)
{
// Make your calculation
cout<<"Would you like to perform other calculation?(Y/N)"<<endl;
cin >> choice;
if(choice =='N'||choice =='n'){
break;
}
}
return 0;
}
But both these solution will consider any character entered exept N/n as "continue", if you want to avoid this:
int main()
{
char choice;
bool run = true;
while(run)
{
// Make your calculation
do{
cout<<"Would you like to perform other calculation?(Y/N)"<<endl;
cin >> choice;
choice = tolower(choice);//Put your letter to its lower case
}while (choice != 'n' && choice != 'y')
if(choice =='n'){
run = false;
}
}
return 0;
}

Answer 2

For some reasons whenever I enter y or Y, the program ends
The return statement is used to return control(and sometimes an optional value), to a functions caller. When return is used inside the called function, the function is terminated. From cppreference.com:
[return] terminates [the] current function and returns [a] specified value to the caller function.
(emphasis mine)
You may be under the impression that the statement return true inside your while-loop is returning the Boolean value of true to your while condition. It is not.
If your end goal is to create a yes/no style program that ends when the user enters "No/no", then you can to make use of the continue and break statements, or use a do/while loop.
Using continue and break
The continue statement is used to immediately skip to the next iteration of a loop, for or while, terminating the current iteration. From cppreference.com:
Causes the remaining portion of the enclosing for, range-for, while or do-while loop body to be skipped.
Used when it is otherwise awkward to ignore the remaining portion of the loop using conditional statements.
(emphasis mine)
The break statement is somewhat similar to continue, but instead of breaking the current iteration and skipping to the next one, it immediately breaks the entire program out of the while loop returning control to the outer-scope. From cppreference.com:
Causes the enclosing for, range-for, while or do-while loop or switch statement to terminate.
Used when it is otherwise awkward to terminate the loop using the condition expression and conditional statements.
(emphasis mine)
After examining the information above, you can modify your program to make use of continue and break, instead of return:
#include <iostream>
using namespace std; // This is for demonstration purposes ONLY
// never use this statement in your actual program. prefix cout and
// cin with std::
int main()
{
char choice;
while(true)
{
cout<<"Would you like to perform other calculations?(Y/N)"<<endl;
cin >> choice;
if(choice == 'Y'|| choice =='y'){
continue; // instead of returning, skip to the next iteration
// and ask again
}else if(choice =='N'||choice =='n'){
break; // return could be used here to break the while loop and
// terminate the program. But be explicit and use a statement specifically
// made for breaking out of loops
}
}
return 0;
}
Using a do/while loop
While the method above would work, I recommend using my second option-- a do/while loop. The do/while has the advantage o being shorter, and not having to make use of any kind of major control flow. From cppreference:
Executes a statement repeatedly, until the value of expression becomes false. The test takes place after each iteration.
If needed you could even add error checking to validate a users input:
#include <iostream>
using namespace std; // This is for demonstration purposes ONLY
// never use this statement in your actual program. prefix cout and
// cin with std::
int main()
{
char choice;
do { // do everything in the do block while...
cout <<"Would you like to perform other calculations?(Y/N)"<< endl;
cin >> choice;
if (choice != 'Y' and choice != 'y' and choice != 'N' and choice != 'n') // if needed add input
cout << choice << " is not a valid option. Try agian" << endl; // validation
} while (choice !='N' && choice !='n'); // the user input does not equal 'N'andr 'n'
return 0;
}
Output
Would you like to perform other calculations?(Y/N)
y
Would you like to perform other calculations?(Y/N)
Y
Would you like to perform other calculations?(Y/N)
n
References and Resources
cppreference.com
cplusplus.com

Answer 3

I just wrote some code and it works like a charm using "goto".
The code also included data validation as well. Many thanks to #Treycos.
int main()
{
char choice ='Y';
bool valid = true;
//some calculation
here:
int x =1+1;
cout<<x<<endl;
while(valid){
cout<<"Would you like to perform other calculation?(Y/N)"<<endl;
cin >> choice;
if(choice =='N' || choice =='n'){
break;
}
if(choice =='Y'||choice =='y'){
goto here;
}
else if(choice != 'N'&& choice != 'n'&&choice != 'Y'&&choice != 'y'){
cout<<"Invalid input."<<endl;
valid = true;
}
}
return 0;
}
Output:
2
Would you like to perform other calculation?(Y/N)
y
2
Would you like to perform other calculation?(Y/N)
Y
2
Would you like to perform other calculation?(Y/N)
$
Invalid input.
Would you like to perform other calculation?(Y/N)
n
Process returned 0 (0x0)
For uppercase n:
2
Would you like to perform other calculation?(Y/N)
N
Process returned 0 (0x0)