While generating double random number 0 is not coming

Question

1.00/5 (3 votes)

See more:

I have written a code to generate random number ranging 0.8 - 1.2
while executing the numbers above 1 coming correctly as 1.2,1.22,1.4 etc

but in the case of 0.xx its coming like .98,.89,etc
I don't need repeating values

tried with zeros but it's repeating the values for other variables also

i'm stuck with this

Need Help

Thank you

What I have tried:

var ans = 'NA';

if (Condition> 0)
        {
          Random random = new Random();
          double val = (random.NextDouble() * (1.2 - 0.8) + 0.8);
          ans= val.ToString("#.##");
        }


        var needed = ans;

this gives .98

Posted 20-Nov-22 16:58pm

Manojkumar-dll

Updated 20-Nov-22 19:29pm

v2

Add a Solution

Comments

Graeme_Grant 20-Nov-22 23:03pm

Your issue is unclear. Please show a specific example.

2 solutions

Solution 1

Do you mean that you want a leading zero '0' for numbers less than 1?

Then you would need:

C#

string ans = val.ToString("0.##");

If you want the same number of decimals, then:

C#

string ans = val.ToString("0.##0");

or...

C#

string ans = val.ToString("N2");

Another way is:

C#

string ans = $"{val:N2}";

Posted 20-Nov-22 17:07pm

Graeme_Grant

Updated 20-Nov-22 17:11pm

v4

Comments

Manojkumar-dll 20-Nov-22 23:17pm

Thank you sir you're right
but there is a error showing as " A local or parameter named 'ans' cannot be declared in this scope because that name is used in an enclosing local scope to define a local or parameter C#"

i think this arises due to ans is declared as variable data type,if that's the problem means how can i solve that

Graeme_Grant 20-Nov-22 23:42pm

I specified the type declaration for the example. How you use it is based on your own code.

Manojkumar-dll 20-Nov-22 23:45pm

yes sir if i use the first example(0.##) means it doesn't exceed 1 but i need values upto 1.2
the sceond one (0.##0) gives like 0.768 but i need only 2 numbers after float

Graeme_Grant 21-Nov-22 0:02am

They are only examples. How you use it is up to you.

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2022-11-20T19:29:00

To add to what Greame says, don't do it like that.
Declare a single Random instance at class level:

C#

private Random random = new Random();

Or:

C#

private static Random random = new Random();

and us that instead. The reason why is simple: when you create a Random instance, it is initialized from the system clock which has a fint=ite resolution and modern processors are well and truly capable of executing many instructions during a single tick of the clock. As a result, if you create a Random instance each time you want to use it, there is a very good chance that you will get the same sequence repeatedly if you execute the same code in a loop.
Creating a single instance means that doesn't happen, unless the random sequence itself actually contains duplicates.

While generating double random number 0 is not coming

2 solutions

Solution 2

Solution 1

Add your solution here

Preview 0

While generating double random number 0 is not coming

2 solutions

Solution 2

Solution 1

Add your solution here

Preview 0

Existing Members

...or Join us