Console.Read() doesn't allow me to input anything the second time

Question

I'm solving a problem about comparing two floating-point numbers and I want to let the user input desired values. So I wrote the following code:

    Console.WriteLine("Enter first number: ");
    double num1 = Console.Read();
    Console.WriteLine("Enter second number: ");
    double num2 = Console.Read();

Unfortunately, I can only input the first number. After the console outputs "Enter first number: " and I enter some number, it simply skips to the end and doesn't let me enter the second number... Any thoughts on that?

Have you seen this: http://stackoverflow.com/questions/6825943/difference-between-console-read-and-console-readline ? — David Tansey, Nov 02 '13 at 23:26
yes I have, but it doesn't matter if I use Read or ReadLine... It's still the same problem. I tried to use ReadLine and parse the value to double, but then it gives me an error message in the console itself... — Boyan Kushlev, Nov 02 '13 at 23:31

score 1 · Accepted Answer · edited May 23 '17 at 12:29

1

That is the default behaviour of Console.Read(). From an answer on Difference between Console.Read() and Console.ReadLine()?

Console.Read() basically reads a character so if you are on a console and you press a key then the console will close. [...]

You should use Console.ReadLine(); instead.

Console.WriteLine("Enter first number: ");
double num1 = double.Parse(Console.ReadLine());
Console.WriteLine("Enter second number: ");
double num2 = double.Parse(Console.ReadLine());

edited May 23 '17 at 12:29

Community

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answered Nov 02 '13 at 23:31

1

It didn't work the first time, but it turned out that when I input a float-point value, I should use a comma, and not a decimal point. In other words, if I input 2.56 it shows me an error message, but when I input 2,56 everything works just fine! Thanks! – Boyan Kushlev Nov 02 '13 at 23:39

geedubb · Answer 2 · 2013-11-02T23:37:02.603

0

Try Console.ReadLine() instead. Console.Read only reads a single character

   Console.WriteLine("Enter first number: ");
   double num1 = double.Parse(Console.ReadLine());
   Console.WriteLine("Enter second number: ");
   double num2 = double.Parse(Console.ReadLine());

Or with TryParse:

   Console.WriteLine("Enter first number: ");
   double num1, num2;
   double.TryParse(Console.ReadLine(), out num1); // double.TryParse() method also returns a bool, so you could flag an error to the user here
   Console.WriteLine("Enter second number: ");
   double.TryParse(Console.ReadLine(), out num2);

edited Nov 02 '13 at 23:37

answered Nov 02 '13 at 23:28

geedubb

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that won't work in this form: `Console.ReadLine()` returns a string. you need to attach a `Convert.ToDouble()` – UnholySheep Nov 02 '13 at 23:29
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we don't speak here about ways of type conversion – AuthorProxy Nov 02 '13 at 23:33

score 0 · Answer 3 · answered Nov 02 '13 at 23:31

It assumes that you enter already \n as a second input. If you enter 2 numbers on the first Read method. Than it tooks 1 number in first read and second number on the second automatically. Just replace with ReadLine() if you want to achieve noraml behaviour,

Console.Read() doesn't allow me to input anything the second time

3 Answers3