how do I validate user input as a double in C++?

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How would I check if the input is really a double?

double x;

while (1) {
    cout << '>';
    if (cin >> x) {
        // valid number
        break;
    } else {
        // not a valid number
        cout << "Invalid Input! Please input a numerical value." << endl;
    }
}
//do other stuff...

The above code infinitely outputs the Invalid Input! statement, so its not prompting for another input. I want to prompt for the input, check if it is legitimate... if its a double, go on... if it is NOT a double, prompt again.

Any ideas?

Try this:

while (1) {
  if (cin >> x) {
      // valid number
      break;
  } else {
      // not a valid number
      cout << "Invalid Input! Please input a numerical value." << endl;
      cin.clear();
      while (cin.get() != '\n') ; // empty loop
  }
}

This basically clears the error state, then reads and discards everything that was entered on the previous line.

validation - how do I validate user input as a double in C++?, Error Checking / Input Validation In C++ (Double) will show you how to validate user input and error check for a double in C++. I hope this  To begin with, inside of the int main() function, create a variable that will store user input. double num; 2. Ask User For Decimal Number. Then, ask the user to enter a decimal number. cout << "Enter a decimal number: "; 3. Receive User Input. Next, we need to receive the users input into the num variable. However, in order to input validate we must place the cin >> num; within the test expression of a while loop. while (!(cin >> num)) { }

failbit will be set after using an extraction operator if there was a parse error, there are a couple simple test functions good and fail you can check. They are exactly the opposite of each other because they handle eofbit differently, but that's not an issue in this example.

Then, you have to clear failbit before trying again.

As casablanca says, you also have to discard the non-numeric data still left in the input buffer.

So:

double x;

while (1) {
    cout << '>';
    cin >> x;
    if (cin.good())
        // valid number
        break;
    } else {
        // not a valid number
        cout << "Invalid Input! Please input a numerical value." << endl;
        cin.clear();
        cin.ignore(100000, '\n');
    }
}
//do other stuff...

Error Checking / Input Validation In C++ (Double), i want know how to validate double from the user's input. #include<iostream> double dInput; cout<<"Enter a decimal value"; cin>>dInput; There are many ways to validate a user input; let's start with a straightforward solution that uses one of the C standard libraries. Stdlib: Scanf The stdlib.h is a header file that imports the C

One way is to check for floating number equality.

double x;

while (1) {
    cout << '>';
    cin >> x;
    if (x != int(x)) {
        // valid number
        break;
    } else {
        // not a valid number
        cout << "Invalid Input! Please input a numerical value." << endl;
    }
}

validation on double and string - C++ Forum, This program error checks a double for valid input. If you like this video, check out these Duration: 0:33 Posted: Dec 5, 2016 Write a program that reads a user's first name and last name from the console input (each on its own line). Both names must not be empty and must consist of two characters or more. If a user inputs an empty string for the first name, you should output "First name can't be empty!

#include <iostream>
#include <string>

bool askForDouble(char const *question, double &ret)
{
        using namespace std;
        while(true)
        {
                cout << question << flush;
                cin >> ret;
                if(cin.good())
                {
                        return true;
                }
                if(cin.eof())
                {
                        return false;
                }
                // (cin.fail() || cin.bad()) is true here
                cin.clear();  // clear state flags
                string dummy;
                cin >> dummy; // discard a word
        }
}

int main()
{
        double x;
        if(askForDouble("Give me a floating point number! ",x))
        {
                std::cout << "The double of it is: " << (x*2) << std::endl;
        } else
        {
                std::cerr << "END OF INPUT" << std::endl;
        }
        return 0;
}

Error Checking numbers (doubles) In C++ / Input Validation #2 , How do I make sure user input is int in C? 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.

bool is_double(double val)
{
bool answer;
double chk;
int double_equl = 0;     
double strdouble = 0.0;
strdouble = val;           
double_equl = (int)val;
chk = double_equl / strdouble;
if (chk == 1.00)
{
 answer = false; // val is integer
 return answer;
} else {
answer = true;  // val is double
return answer;
}
}

Check if input is integer type in C, How do you check if the input is an integer in C++? This is extremely important because , an unhandled wrong input might have the complete ability to crash a system. C++ has some good validation techniques that can be used to validate most kind of inputs. This post discusses some of the techniques and its shortcomings and what could be done to improve the quality of validation.

Validating user input in C++, . clear() - This is used to clear the error state of the buffer so that further processing of input can take place. Verifying User Input as a Double . Verifying User Input as a Double and I need a way to validate that the data entered is an int for every iteration of the loop

Validate user input, C++ has some good validation techniques that can be used to validate most kind of inputs. This post discusses some of the techniques and its shortcomings and  1) If the user inputs more than one character, reject the input and prompt the user for a new input, or 2) Just take the first character the user inputs and discard whatever else the user might have entered after that (if anything). For #2, you can just do

Validate Input as a Int/Double, Learn about Validate user input. If sscanf fails the input is not valid: ( void ) { char buff[BUFSIZ]; char s[7] = {0}; /* +1 for nul */ double f; int n;  Java User Input. The Scanner class is used to get user input, and it is found in the java.util package. To use the Scanner class, create an object of the class and use any of the available methods found in the Scanner class documentation. In our example, we will use the nextLine() method, which is used to read Strings:

Comments
  • Be sure to look at the formatting guide next time you make a post. Thanks.
  • what do you mean? what is wrong with the formatting?
  • I edited your post, but before the edit the code wasn't formatted. Click "edited N mins ago" to see the original post.
  • I see... I thought I had done that... must have missed the formatting button and pressed something else by accident. My apologies.
  • is this inside the while(1) loop?
  • Instead of while(cin.get() != '\n');, I'd just use cin.sync();. It's more readable in my opinion.
  • In UNIX, it's a \r\n, so there's still a \n at the end. So yes, it will work.
  • @chaosTechnician, cin.sync() doesn't work for me when I try this example. the while(cin.get() != '\n'); does though. Any idea why?
  • No, it's written before: cin.sync(); Here's a snippet of the code: int main(int argc, char *argv[]) { double x; while (1) { cout << '>'; if (cin >> x) { break; } else { // not a valid number cout << "Invalid Input! Please input a numerical value." << endl; cin.clear(); cin.sync(); // while(cin.get() != '\n'); } } cout << "you entered: " << x << endl; return 0; }
  • this is still causing the infinite loop to print the Invalid Input!... its not prompting for another input.
  • What does the cin.ignore() do?
  • Same thing as casablanca's loop, it throws away all characters up to and including the newline to get rid of whatever non-numeric data caused the first extraction to fail.
  • how does this check for type double?
  • This just checks if the input had a fractional part, which is probably not what was meant by the question.