vb results are rounded to two decimal places.

vb data rounding to two decimal places

VB

There are several ways to round, for example for a=123.4567, the requirement is to retain two decimals according to the principle of rounding, i.e., the result of a is 123.46

There are several ways:

a=round(a

,2)

a=format(a, “##.00”)

a=int(a*100+0.5)/100

In VB if x is a real variable what function should be used to make its output retain two decimals?

In VB if x is a real variable, to make its output retain two decimals apply

Round(x,2) function, it will be the existing excess number of decimal places to achieve “rounding”.

What is the function in vb that takes the value (2 decimal places)?

VB has functions dedicated to rounding

Round function: returns the value rounded to the specified number of places.

Syntax: Round (expression[,numdecimalplaces])

Parameters: expression (mandatory) – the numerical expression to be rounded; numdecimalplaces (optional). number indicates how many places to the right of the decimal point for rounding, if omitted, the Round function returns an integer.

The following example uses the Round function to round values to two decimal places:

DimMyVar,pi

pi=3.14159

MyVar=Round(pi,2)’MyVarcontains3.14.

VB rounding to retain two decimals, how is this done?

Round is not rounding, it’s “close rounding”. It returns the integer closest to the argument. If the argument is the median of two integers, one of which is even and the other odd, then an even number is returned.

The behavior of this method follows section 4 of IEEE Std 754. This rounding is sometimes called proximity rounding or banker’s rounding.

VB rounding problem, urgent!

The problem is int(XXX*100)/100

This sentence means to keep two decimals, you can replace it with int(XXX*10)/10

Excuse me, excuse me, and forgot one more thing

Please see the following:

Int(N*100+0.5)/100 which is A formula that preserves two decimal places

So you want to preserve one decimal place

Int(N*10+0.5)/10

Verify that the value of 1.76 is substituted

1.76 multiplied by 10 yields 17.6

17.6 plus 0.5 yields 18.1

18.1 rounded up yields 18

< p>18 divided by 10 gives 1.8

Correct!