Verifying User Input as Integer: User has to enter input twice

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I am prompting the user to enter an integer value. When the value is incorrect, the program works. However, when the user enters an integer input, the user needs to enter the input twice.

I looked at other tutorials on how to use the while loop to catch erroneous input, and that part worked for me. However, the integer values need to be entered twice in order for the program to run.

#include <iostream>
using namespace std;

int main() {
    cout << "*************************************************" << endl;
    cout << "******************|DVD Library|******************" << endl;
    cout << "*************************************************" << endl;
    cout << "1.\tAdd DVD" << endl;
    cout << "2.\tDelete DVD" << endl;
    cout << "3.\tSearch DVD" << endl;
    cout << "4.\tList All DVDs in the Library" << endl;
    cout << "5.\tAdd DVD to Favorites List" << endl;
    cout << "6.\tDelete DVD from Favorites List" << endl;
    cout << "7.\tSearch DVD in Favorites List" << endl;
    cout << "8.\tList All DVDs in Favorites List" << endl;
    cout << "9.\tQuit" << endl;
    cout << "*************************************************" << endl;

    int input;
    cin >> input;
    while (!(cin >> input)) {
        cin.clear();
        while (cin.get() != '\n')
            continue;
        cout << "Please enter an integer --> " << flush;
    }
    if (input < 1 || input > 9) {
        cout << "Invalid input! Please try again!" << endl;
    }
    return 0;
}

You ask for the input twice:

cin >> input;
while(!(cin >> input )){

Removing the first line might make it work you intended.

Basic Input-Output - Programming in Python, Input usually means the data entered by the end-user of the program. Of course for programs with graphical user interfaces, the output is much more complex Using a type converter, one could get an integer in string form from the user, Once you can check that the input is indeed a whole number, use the code twice to� There are many things that can "leave stuff pending". For example, a call to nextInt() will leave all of the input pending if the input cannot be interpreted as an int If the user types "4 5", the Scanner will read the 4 and leave the 5 pending for the next read It is probably best to first use hasNextnt() to ensure that an int can be read.

'The user has to enter the input twice' Look at your code

int input;
cin >> input;
while(!(cin >> input )){

How many times do you ask the user for input?

You'd have more luck with this

int input;
while(!(cin >> input )){

Your error recovery code looks reasonable, haven't tested it though.

How to stop user from entering same number twice?, i need to create a program that will accept 5 numerical inputs and determine the highest, and i also need to stop user from using the same number twice and do You can use set<int> and check the std::set's insert signature if� To use them as integers you will need to convert the user input into an integer using the int() function. e.g. This line of code would work fine as long as the user enters an integer. If, by mistake, they enter letters or punctuation signs, the conversion into an integer will fail and generate an exception (error) and the program would stop

int input;
while (cout << "Your choice: ",
       !(cin >> input) || input < 1 || 9 < input)
{
    cin.clear();
    while (cin.get() != '\n');
    cerr << "Invalid input! Please try again!\n";
}

C Basics - C Programming Tutorial, For example, a int variable can store an integer value such as 123 , but NOT real number 2.0; // Use average, mark1 and mark2 int mark1; // Error: Declare twice mark2 In C, you can use scanf() function of <stdio.h> to read inputs from keyboard. Checking the return code of scanf() is recommended for secure coding. 14 thoughts on “ How to validate numeric-integer input in C ” Chuck March 10, 2013 at 3:42 pm. Instead of checking for what you could do is read in that character and not use it.

Thanks everyone! The "cin >> input;" line was unnecessary. At first, I left it there because it would actually tell the user the error message if the user entered a numeric input such as a double. So, if the user entered something like 3.3, the program would display an error message that I specified ("Please enter an integer" line). However, the program in this case (when there is a double) asks the user to prompt for the integer input twice and then continues the program. When I delete the said unnecessary line, the program accepts a double input, but what it does, it takes the numeric value before the decimal point and uses it as the integer. So, a value of 1.2 is recorded as 1 when I tested it. I'm unsure why this phenomenon happens, but the program works otherwise. Maybe it accounts for human error?

#include <iostream>
using namespace std;

int main() {
    cout << "*************************************************" << endl;
    cout << "******************|DVD Library|******************" << endl;
    cout << "*************************************************" << endl;
    cout << "1.\tAdd DVD" << endl;
    cout << "2.\tDelete DVD" << endl;
    cout << "3.\tSearch DVD" << endl;
    cout << "4.\tList All DVDs in the Library" << endl;
    cout << "5.\tAdd DVD to Favorites List" << endl;
    cout << "6.\tDelete DVD from Favorites List" << endl;
    cout << "7.\tSearch DVD in Favorites List" << endl;
    cout << "8.\tList All DVDs in Favorites List" << endl;
    cout << "9.\tQuit" << endl;
    cout << "*************************************************" << endl;

    int input;
    while (!(cin >> input)) {
        cin.clear();
        while (cin.get() != '\n')
            continue;
        cout << "Please enter an integer --> " << flush;
    }
    if (input < 1 || input > 9) {
    cout << "Invalid input! Please try again!" << endl;
}
return 0;

}

An Introduction to C Programming for First-time Programmers, Enter first integer: 55 Enter second integer: 66 The sum of 55 and 66 is 121. We then use the scanf() function to read the user input from the keyboard and You can then use a conditional statement to check whether the number is odd or � Please enter a positive number: qwerty "qwerty" is not a valid number. @@@ "@@@" is not a valid number. -100 Please enter a positive number: 99 You have entered a positive number 99. Another example is to validate if user correctly input letters to guest a secret word.

While Loops and Input, They set up an infinite loop that runs until the user does something to end the the kind of input you are looking for, and then it waits for the user to enter a value. process them. current_user = unconfirmed_users.pop() print("Confirming user � Get User Input. You have already learned that Console.WriteLine() is used to output (print) values. Now we will use Console.ReadLine() to get user input.. In the following example, the user can input his or hers username, which is stored in the variable userName.

Enter your number 29 User input is Number. Note: isdigit() function will work only for positive integer numbers. i.e., if you pass any float number, it will say it is a string. Let’s execute the above program again to validate it. Output two: Enter your number 22.40 User input is string. So it is better to use the first approach.

Keep in mind that if the user doesn't actually enter an integer then this code will throw an exception, even if the entered string is a floating point number. Input Exception Handling There are several ways to ensure that the user enters valid information.

Comments
  • . o O ( DVDs are so '90s )