int.Parse not working

Question

static void Main()
{        
    Console.Write("Please input a number: ");
    Console.WriteLine("\n The number you selected was {0} \n", method());
}

static int method()
{
    int var = int.Parse(Console.ReadLine());
    return var;
}

The above code throws a format exception. I tried storing the input in a string variable and then parsing, but it had the same problem. I also tried using the Convert class and still had the same problem. I would appreciate if someone could show me where I am wrong.

I am trying to convert 23.4, for example. (It works for Natural numbers but why not 4345.5, for example)

Because int.Parse() will parse an integer number, if you write 4345.5 it'll throw an exception because it's not integer. Use float.Parse() or (better) int.TryParse/float.TryParse — Adriano Repetti
– Adriano Repetti, Commented Oct 27, 2014 at 16:23
@StefanosVakirtzis see my answer. You still need the float part but you can return only the integer part. 3.6 becomes 3, for example. — mihai
– mihai, Commented Oct 27, 2014 at 16:38
@StefanosVakirtzis: you can use the code you've accepted to get the double. Then use (int)number to get an integer where the decimal part is simply truncated. If you want that it gets rounded use Math.Round first. — Tim Schmelter
– Tim Schmelter, Commented Oct 27, 2014 at 16:38

Tim Schmelter · Accepted Answer · 2014-10-27 16:46:02Z

23.4 is not an integer, so you cannot use int.Parse (or int.TryParse). Instead you have to parse it to a decimal number like decimal or double. You can use the TryParse methods like Double.TryParse to prevent an exception if it's not a valid number:

string input = Console.ReadLine();
double number;
if(double.TryParse(input.Trim(), out number))
{
    // valid number
    Console.WriteLine("\n The number you selected was {0} \n", number);
}
else
{
     Console.WriteLine("Please enter a real number.");
}

Update from your comments i can see that you want to display an integer.

You can use (int)number to get an integer where the decimal part is simply truncated. If you want that it gets rounded use Math.Round first.

int integer = (int) number; // decimal part is truncated
integer = (int) Math.Round(number, 2, MidpointRounding.AwayFromZero); // rounded to two digits

If you just want to display a string without the decimal part you can also use format strings.

string numberString = number.ToString("N0");

Stuart Carnie · Accepted Answer · 2014-10-27 16:24:32Z

1

You should be using double or float for rational numbers:

float num = float.Parse(Console.ReadLine())

answered Oct 27, 2014 at 16:24

Stuart Carnie

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Comments

mihai · Accepted Answer · 2014-10-27 16:35:59Z

1

You can return the integer part of your input string (3.6 becomes 3, for example) like this:

static int method()
{
    float var = float.Parse(Console.ReadLine());
    return (int)Math.Floor(var);
}

answered Oct 27, 2014 at 16:35

mihai

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Comments

Servy · Accepted Answer · 2014-10-27 16:23:53Z

0

Integer32.Parse is for parsing 32 bit integers. It cannot parse numbers that cannot be represented as a 32 bit integer, such as 23.4, and it will throw an exception when it cannot parse the data. Use a different numeric representation if you wish to represent non-integer numbers.

answered Oct 27, 2014 at 16:23

Servy

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