Please Help ----- Getting wrong result in C# than VB.Net --- Urgent

[pargat.singh@gmail.com]

Hi Everyone:

Please help as i need to write some rounding function in C#. i
Wrote one but does not give me correct result. Can some one please
correct me

C# Codes

############### ############### ############### ############### #############
dblVal = 1234567.89

private int ThreeSifFig(dou ble dblVal)
{
int lngvalue;
int threesigfig;
int lngvaluelen;

lngvalue = (int)(Math.Abs( dblVal));
lngvaluelen = lngvalue.ToStri ng().Length;
if ((lngvaluelen <= 1))
{
threesigfig = 0;
}
else if ((lngvaluelen <= 3))
{
threesigfig = (lngvalue / 10) * 10;
}
else
{
threesigfig = (lngvalue / 10 ^ (lngvaluelen - 3) * 10) ^
(lngvaluelen - 3);
}
if ((dblVal < 0))
{
threesigfig = (threesigfig * -1);
}

return threesigfig;

}

Return Result is 123500 but the expected result is 1230000

############### ############### ############### ############### ###############

VB.NET code which return correct result
Protected Sub Page_Load(ByVal sender As Object, ByVal e As
System.EventArg s) Handles Me.Load

Dim sval As Double = 1234567.89
Dim lngvalue, lngvaluelen As Long
Dim threesigfig As Long
lngvalue = Int(Math.Abs(Va l(sval)))
lngvaluelen = Len(CStr(lngval ue))
If lngvaluelen <= 1 Then
threesigfig = 0
ElseIf lngvaluelen <= 3 Then
threesigfig = Int(lngvalue / 10) * 10
Else
threesigfig = Int(lngvalue / 10 ^ (lngvaluelen - 3)) * 10
^ (lngvaluelen - 3)
End If
If sval < 0 Then
threesigfig = threesigfig * -1
End If
End Sub

Result is 1230000 [Correct]

Thanks in advance.

[=?Utf-8?B?RmFtaWx5IFRyZWUgTWlrZQ==?=]

RE: Please Help ----- Getting wrong result in C# than VB.Net --- Urgen

That is a very complicated way to find the value you are after, and,
it is not generic. If you wanted to find the value to six digits, you would
need a new function.

You should find the integer log base 10 value for the number. So for your
value of 1234567.89, you have a log base 10 equal to 6. Now if you want
three significant digits, subtract two (or one less than the sig digits) off
the
value of six. This gives you four. Raise 10 to the 4th and you have 10000.

Divide your value by the power of ten just computed (1234567.89 / 10000)
and get the integer value (123). Now multiply the result by the 10000.

Result = 1230000.

Hope this helps.


"pargat.singh@g mail.com" wrote:

Hi Everyone:
>
Please help as i need to write some rounding function in C#. i
Wrote one but does not give me correct result. Can some one please
correct me
>
C# Codes
>
############### ############### ############### ############### #############
dblVal = 1234567.89
>
private int ThreeSifFig(dou ble dblVal)
{
int lngvalue;
int threesigfig;
int lngvaluelen;
>
lngvalue = (int)(Math.Abs( dblVal));
lngvaluelen = lngvalue.ToStri ng().Length;
if ((lngvaluelen <= 1))
{
threesigfig = 0;
}
else if ((lngvaluelen <= 3))
{
threesigfig = (lngvalue / 10) * 10;
}
else
{
threesigfig = (lngvalue / 10 ^ (lngvaluelen - 3) * 10) ^
(lngvaluelen - 3);
}
if ((dblVal < 0))
{
threesigfig = (threesigfig * -1);
}
>
return threesigfig;
>
}
>
Return Result is 123500 but the expected result is 1230000
>
############### ############### ############### ############### ###############
>
VB.NET code which return correct result
Protected Sub Page_Load(ByVal sender As Object, ByVal e As
System.EventArg s) Handles Me.Load
>
Dim sval As Double = 1234567.89
Dim lngvalue, lngvaluelen As Long
Dim threesigfig As Long
lngvalue = Int(Math.Abs(Va l(sval)))
lngvaluelen = Len(CStr(lngval ue))
If lngvaluelen <= 1 Then
threesigfig = 0
ElseIf lngvaluelen <= 3 Then
threesigfig = Int(lngvalue / 10) * 10
Else
threesigfig = Int(lngvalue / 10 ^ (lngvaluelen - 3)) * 10
^ (lngvaluelen - 3)
End If
If sval < 0 Then
threesigfig = threesigfig * -1
End If
End Sub
>
Result is 1230000 [Correct]
>
Thanks in advance.
>

[=?Utf-8?B?RmFtaWx5IFRyZWUgTWlrZQ==?=]

RE: Please Help ----- Getting wrong result in C# than VB.Net --- Urgen

threesigfig = (lngvalue / 10 ^ (lngvaluelen - 3) * 10) ^

(lngvaluelen - 3);
>

The above line does not represent a translation from the VB code. Look, for
example, at the result of 10 ^ 4 versus what you think it should be.