INT |
INT(number) |
Returns the integer portion of a number (data type). |
number | The number (data type) |
REMARKS |
* The fractional/decimal part of the number is removed. * This function is identical to the FIX function for positive numbers. * This function truncates without rounding. * The value returned has the same data type as the value passed in. * If "number" is Null, Null is returned. * If "number" < 0, then the first negative integer less than or equal to "number" is returned. * You can use the FIX function to return the integer portion of a number. * You can use the ROUND function to return a number rounded to a given number of decimal places. * The equivalent .NET function is Microsoft.VisualBasic.Conversion.Int * For the Microsoft documentation refer to learn.microsoft.com |
Debug.Print Int(-8.9) '= -9
Debug.Print Int(-8.4) '= -9
Debug.Print Int(-8.1) '= -9
Debug.Print Int(-7.9) '= -8
Debug.Print Int(-1.9) '= -2
Debug.Print Int(-1.1) '= -2
Debug.Print Int(-0.1) '= -1
Debug.Print Int(0.2) '= 0
Debug.Print Int(0.5) '= 0
Debug.Print Int(0.6) '= 0
Debug.Print Int(0.9) '= 0
Debug.Print Int(1.1) '= 1
Debug.Print Int(1.5) '= 1
Debug.Print Int(12.01) '= 12
Debug.Print Int(12.34) '= 12
Debug.Print Int(12.99) '= 12
Dim iInteger As Integer
iInteger = 20
Debug.Print TypeName(Int(iInteger)) '= Integer
Dim lLong As Long
lLong = 20.5
Debug.Print TypeName(Int(lLong)) '= Long
Dim dbDouble As Double
dbDouble = 20.5
Debug.Print TypeName(Int(dbDouble)) '= Double
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