CODEWITHSUNDEEP
vb6
PLAA Hemantha
PLAA Hemantha

Reputation: 75

Fix operator problems in VB6

All, I'm intermediate for VB6. I want to get remainder from divided two doubles. I used ,

Dim a As Double, b As Double, result As Double
b = 8333.33
a = 58333.31

result = a - (b * Fix(a / b))

result should be 0. But it is not. a/b =7 and no remain. So Fix(a / b) should be 7. But Fix(a / b)=6, Why?

Upvotes: 3

Views: 574

Answers (2)

Hel O'Ween
Hel O'Ween

Reputation: 1486

BTW, another trick to get it working - depending on the floating point numbers you need to work with - is to use Currency instead of Double

Dim a As Currency, b As Currency

b = 8333.33
a = 58333.31

Debug.Print Fix(a / b) ' Returns 7
Debug.Print a - (b * Fix(a / b)) ' Returns 0

Upvotes: 3

StayOnTarget
StayOnTarget

Reputation: 13048

If you use this as a workaround I think you will get the result you expect:

Dim a As Double, b As Double, intermediate As Double, fixed1 As Double, fixed2 As Double

b = 8333.33
a = 58333.31

intermediate = a / b
fixed1 = Fix(intermediate)
fixed2 = Fix(a / b)

Debug.Print intermediate ' 7
Debug.Print fixed1       ' 7
Debug.Print fixed2       ' 6

Documentation for Fix says:

Int, Fix functions

Returns the integer portion of a number.

... Remarks Both Int and Fix remove the fractional part of number and return the resulting integer value.

(that is from VBA docs but should be the same for VB6).

Apparently when VB evaluates a / b within the context of the function call it results in a floating point value very slightly < 7.

Upvotes: 2

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