[resolved] retrieve the integer value after dividing by a number

[danzey]

hello i would like a code to retrieve the integer value
example after dividing a number , i got 100,6 and i would like 100 (or 101 if it is possible)
it 's to integrate in numericupdown.value
thanks for your answer

[PlausiblyDamp]

https://msdn.microsoft.com/en-us/lib...v=vs.110).aspx is probably what you are after.

[danzey]

i am a beginner i found round methode but i can't use it.Please help me with an example.

[.paul.]

Try:

intValue = Cint(Math.Floor(10/3))

[.paul.]

Here's a rounding example:

'to get the integer value and discard the fractional part
Dim intValue As Integer = CInt(Math.Floor(10 / 3))
MsgBox(intValue)
'to use banker's rounding
intValue = CInt(Math.Round((10 / 4), MidpointRounding.AwayFromZero))
MsgBox(intValue)

[danzey]

Try:

intValue = Cint(Math.Floor(10/3))

so resuming,
i have this addition: numericupdown3.value = numericupdown1.value+ (Cint(Math.Floor((numericupdown2.value) /2))

am i right?

[.paul.]

That code will always round down. If you want midpoint rounding, see my second example...

[Shaggy Hiker]

Another thing you can do is use the integer divide \, rather than the floating point divide /. This does the round down, too, so you can write:

numericupdown3.value = numericupdown1.value + numericupdown2.value \ 2

That gives you the same result, since it will also always round down.

[danzey]

thank you very much! it works!