C++ Phone Number Program - c++

Question

I am new to this forum and am looking for some assistance, any help would be greatly appreciated! I am stuck on this assignment for my programming fundamentals 1 class and am desperate at this point as I have been stuck for hours on end. Thank you.
Here is the prompt:
Many websites ask for phone numbers. The problem is that there are so many
different ways to represent a phone number. Examples include 817-555-1234,
817 555 1234 (c), and [phone] x23. Write a C++ program which inputs
a string containing a phone number in any format and outputs it in the standard
format. For this assignment, the standard format is [phone].
Your c++ program should:
1. Input a string including the number
2. Copy only the digits from the input string into another string
3. Issue an error message if the input string does not contain exactly 10
digits
4. Output the phone number in standard format
#include "stdafx.h"
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <string>
#include <cctype>
using namespace std;
const int NUM_LENGTH = 10;
string ReadAndValidateUserNumber(string userNumber);
int main()
{
string userNumber;
ReadAndValidateUserNumber(userNumber);
system("PAUSE");
return 0;
}
string ReadAndValidateUserNumber(string userNumber)
{
bool check = false;
while (!check)
{
check = true;
cout << "Please enter a Number: ";
cin >> userNumber;
if (userNumber.length() != NUM_LENGTH)
cout << "The phone number may contain 10 digits only. \n";
else
{
userNumber.insert(0, "(");
userNumber.insert(4, ")");
userNumber.insert(8, "-");
for (int i = 0; i < userNumber.length(); i++)
{
if (isdigit(userNumber[i]))
{
userNumber = NUM_LENGTH;
}
}
}
if (!check)
{
cout << "Invalid Entry! Please try again." << endl;
}
}
return userNumber;
}

Answer 1

You appear to have a logic error.
bool check = false;
while (!check)
{
check = true;
....
if (!check)
{
cout << "Invalid Entry! Please try again." << endl;
}
}
You'll note check never becomes false again. Hence the bottom if statement will never execute and the loop will never iterate more than once as check is always true. You probably want to make
check = true
conditional on the correct format being entered.

Answer 2

The code is in a horrible state. Calm down and focus. There are many problems aside from what you've written. I recommend you to carry the working code to any code reviewing site.
You're doing the check incorrectly, as well as what you're doing after the check is wrong as well.
userNumber = NUM_LENGTH;
This will truncate NUM_LENGTH into char, so you'll have something strange inside userNumber.
This is the right check:
if (!isdigit(userNumber[i]))
{
std::cout << "input should be only numbers\n";
return; //function should return void
}
^^ this is what you should do if you want to break on non number input, if you want to skip non digits, you can use std::copy_if into another string, or tweak the above thing to do it for you. You can also use std::remove_if if you want to remove the non number stuff.
std::string fetched_input;
std::copy_if(userNumber.begin(), userNumber.end(), fetched_input.begin(), std::isdigit);
^^ this will give you the string only with numbers.
Also, you're not doing the check in the correct place. Think about it again.