How while(!(cin >> x)) works to re-prompt for input - c++

Question

while(!(cin >> ar[i]))
{
cin.clear(); // clears bad input
while(cin.get() != '\n')
continue;
cout << "Invalid input, please enter valid scores";
}
The above code is from a much larger file. I copied this bit of a code from one of my textbooks and I don't really feel comfortable using it as I do not understand how this works.
I am using it as a measure to handle input errors.
So ar is an empty array of integers, if I decide to enter 'k', then
!(cin >> ar[i])
is true.
From here I clear the input buffer (I think that is correct, I'd like someone to confirm or dispute this please). The terminal then prints "Invalid input..."
Now if I just press Enter nothing happens, but isn't Enter the newline char? So shouldn't the code read
while(cin.get() == '\n'
?

Answer 1

while(!(cin >> ar[i]))
This tries to parse a value from cin and store it in ar[i]. By default, >> skips whitespace first, then sees if the characters in cin describe a legal value for whatever the type of ar[i] is. If a legal value is found, then cin stream state remains good, and its operator bool() const will kick in given the boolean not/! operation, such that the while loop will break.
If parsing fails though, the stream state will be set to one or more of:
bad (if there's some unrecoverable stream error, like stdin supplied over a network connection that gets disconnected),
fail (if the characters just didn't form a legal value for the type), or
eof (end of file, for a "proper" shutdown/close of the input, as supplied by ^D in UNIX/Linux, ^Z in Windows, and the end of input when a program's invoked as in echo input | program).
All the above modalities are described under "State Functions" here.
If the loop is entered due to any of the error conditions above...
{
cin.clear(); // clears bad input
...this does NOT clear any input data from the stream, but does clear the bad, eof and fail state flags, after which further input attempts can be made, though a stream that was in bad or eof state is likely to immediately reenter that state when further input is attempted (but not always - some OS may allow successful input after an eof conditions for std::cin if the user types/generates an EOF code then types actual text again...
while(cin.get() != '\n')
continue;
This tries to read characters from the terminal until a newline \n is encountered. The idea's clearly to clear out the rest of the presumed unparse-able input that might have led to a fail condition earlier. Sadly, if the problem was, or becomes, a bad or eof condition then this loop will hang the program, spinning burning CPU to no avail.
cout << "Invalid input, please enter valid scores";
}
If the problem was simply a mistyped value and no bad or eof condition, then the cout will prompt for further input.
Now if I just press enter nothing happens, but isnt enter the newline char?
Whenever the outer loop is executing cin >> ar[i] it will skip whitespace, including any extra newlines you type, until it sees some input (which may need to be a full newline-terminated line to get flushed by the terminal or program feeding it to the program), or a bad or eof condition. The inner while-loop is not there to get rid of empty lines - it's trying to discard the line with presumed non-numeric text in it.
Corrected code:
while (!(std::cin >> ar[i]))
{
if (std::cin.bad() || std::cin.eof())
{
std::cerr << "Fatal error on cin while reading numbers\n";
exit(EXIT_FAILURE);
}
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "Invalid input, please enter valid scores\n: ";
}

Answer 2

The goal of this entire code is to keep printing errors and requesting input until the user enters a valid number before continue further into the program.
The loop that the entire code is enclosed in...
while(!(cin >> ar[i]))
says to loop if the input in the cin stream is invalid. Since ar is an array of integers, the input would be invalid if it is not a number.
cin.clear(); // clears bad input
When the cin stream encounters invalid input, the program begins executing the loop and continues to this line of code. Since the stream encountered invalid input, it has a flag that says there is an error. This requires that you, as the comment puts it, "clear bad input." Basically what this does is get rid of this flag. If this is not done, the flag will remain in the stream and the program will encounter another error next time the cin stream is used, regardless of whether or not the user input is valid.
while(cin.get() != '\n')
continue;
When the program took the input from the user, it took the string, char, whatever was the invalid input, but left the '\n' in the cin stream. To prevent errors the next time the cin stream is used, the program must get rid of that '\n', along with anything else that was lefty behind. cin.get() reads only one char from the cin input stream and returns it. Since the return value of this function is not being assigned to anything, all this does is discard the char that was read. In the case that more than the '\n' was left behind in the stream, this loop checks the value of the read char before disposing of it. This way, the loop keeps executing while the char that is read is NOT '\n'. Or in other words, it loops until the char that is read IS '\n'. The only thing that continue does, is tell the program to skip the rest of the current iteration of the loop and start the next iteration. In this situation, this is equivalent to giving the loop an empty body. You could very well replace continue; with {} and the program would do the exact same thing.
cout << "Invalid input, please enter valid scores";
The program has cleared the flag in cin and has also cleared any data from cin that may have been left behind. Now that all the errors from the invalid input have been handled, there is only one more thing to do. It's time to notify he user that the input was invalid and request new input. If the user enters more invalid input, the loop repeats. If the user enters valid input, the program continues on to the next line of code.