Command Line Argument & int.TryParse()

12y1.7kVisual Basic .NET

This following program will check whether value is provided or not in the Command Line Argument, if value is provided whether it is a numeric or not, if it is a numeric whether it is an odd or even.

I have a problem in how trueORfalse = int.TryParse(args[0], out n) behave.

We know inside the if condition if the expression is true that statement becomes o/p.

But in the program if a Command Line Argument is non-numeric though trueORfalse expression is false (Step Into (F11-key) will show the result) statement becomes o/p. How can this situation be explained? Problem is highlighted.

using System;

class Program
{
static void Main(string[] args)
{
int n ;
bool trueORfalse = false;

if(args.Length==0)
Console.WriteLine("Command Line Argument is not provided");
else
if (trueORfalse == int.TryParse(args[0], out n))
Console.WriteLine("Provid a numeric value");
else
if(n%2==0)
Console.WriteLine("Even");
else
Console.WriteLine("Odd");

Console.Read();
}
}

7 Replies

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12y ago
Thank you for your explanation.
0
Vulpes12y ago
You'll find this code easier to understand if you add some more curly brackets and indentation:

int n;

if (args.Length == 0)
{
Console.WriteLine("Command Line Argument is not provided");
}
else if (int.TryParse(args[0], out n))//NOT operator has been removed
{
if (n % 2 == 0)
Console.WriteLine("Even");
else
Console.WriteLine("Odd");
}
else
{
Console.WriteLine("Provide a numeric value");
}

It's now easier to see that the 'else if' clause is only entered if the command line argument is numeric.

Otherwise the code in the final 'else' clause is executed.
0
12y ago
Now NOT operator has been removed and Console.WriteLine("Provid a numeric value") code has been moved to the bottom. But there is no change in the o/p how this can be explained. Problem is highlighted.

using System;

class Program
{
static void Main(string[] args)
{
int n;

if (args.Length == 0)
Console.WriteLine("Command Line Argument is not provided");
else
if (int.TryParse(args[0], out n))//NOT operator has been removed

if (n % 2 == 0)
Console.WriteLine("Even");
else
Console.WriteLine("Odd");
else
Console.WriteLine("Provide a numeric value");

Console.Read();
}
}
0
12y ago
Now I can understand. Thank you for your explanation.
0
Vulpes12y ago
Well, normally, instead of this line:

if (trueORfalse == int.TryParse(args[0], out n))

you'd simply write:

if (!int.TryParse(args[0], out n))

The confusing thing about the code you've posted is that the return value of int.TryParse method is being compared with the value of the bool variable, trueORfalse, which has already been initialized to false.

So, there's no contradiction here - it's just that the method is being used in a way which you're not used to seeing.

0
12y ago
Even though the expression in if (trueORfalse == int.TryParse(args[0], out n)) is evaluated false the statement belongs to that is executed.

Normally the expression in if evaluated true only the statement belongs to that will be executed. How above contradiction can be explained.
0
Vulpes12y ago
Well, as you probably know, int.TryParse attempts to parse its first (string) argument to an integer.

If it succeeds, the method returns true and the integer is returned in the second (output) parameter.

It is fails, the method returns false and a value of 0 is returned in the second parameter.

So, here, if a command line argument is provided, an attempt is made to parse it and if the attempt fails (i.e. the return value equals trueORfalse which is false), the user is asked to provide a numeric value. Otherwise the program goes on to print whether the number is odd or even.
0