C# double works differntly for the same set of values

[B MANOHARAN]

Hi
Can you help me in understanding the following.

double x,y;
x=2.5;
y=22/7*x*x;;

Here the value displayed is 18.75.

when the above code is written as given below the value is 19.642857

x=2.5
y=x*x*22/7;

Actual value shall be 19.642857 only why it gives 18.75 in the first case
as both are literally same.

BMano

[Plater]

Very strange.
Look at this:

double y;
y = 22 / 7;
Console.WriteLine(y.ToString()); 
 

The output is "3".

Using two integers (despite that they also fit as doubles) tells the compiler to do integer math. So append a .0 to your numbers:

double y;
y = 22.0 / 7.0;
Console.WriteLine(y.ToString()); 
 

Doing that gave me the 19 number each time. I think only ONE hardcoded (like the 22 or the 7) number has to be #.0 for it to work correctly.

Alternatively, type-casting one of your integer values toa double will work as well:
y = (double)22 / 7 * x * x; //gives the 19.whatever number

[B MANOHARAN]

Dear Plater
Thank you very much for your reply.

but i need a bit more clarification. double x=2.5; double y;
As yo have mentioned y=22.0/7.0*x*x gives right value.
when it is written as y=x*x*22/7 also gives right value, only
y=22/7*x*x gives wrong value.

does it means that in y=x*x*22/7 the first two value multiplied gets a value of double and hence subsequent multiplying by 22 and dividing by 7 gives right value.
And in the case y=22/7*x*x the first division takes place as integer and subsequently multiplied by x (2..5) and x which gave 18.5 ( a double value.)

BMano

[r035198x]

Dear Plater
Thank you very much for your reply.

but i need a bit more clarification. double x=2.5; double y;
As yo have mentioned y=22.0/7.0*x*x gives right value.
when it is written as y=x*x*22/7 also gives right value, only
y=22/7*x*x gives wrong value.

does it means that in y=x*x*22/7 the first two value multiplied gets a value of double and hence subsequent multiplying by 22 and dividing by 7 gives right value.
And in the case y=22/7*x*x the first division takes place as integer and subsequently multiplied by x (2..5) and x which gave 18.5 ( a double value.)

BMano

That is correct. It's the division part that's important here. When the divisor and the dividend are integers, the result is truncated but when one of them is a double then the value of the expression becomes double.

[Plater]

you gotta think this:
double x; //x is a double
22; //22 is only an integer