Conditional function that returns one of two values.
Syntax
IIf ( condition, expr_if_true, expr_if_false )
Parameters
condition
The condition to test.
A non-zero value evaluates as true, while a value of zero evaluates as false.
expr_if_true
An expression to evaluate and return if condition is true.
It must return:
- a numeric value, which can be an integer, floating point number or a pointer, including Boolean,
- or a string value,
- or an UDT value.
expr_if_false
An expression to evaluate and return if condition is false.
It must be same type as expr_if_true (either numeric, either string or UDT).
Return Value
if condition is non-zero, expr_if_true, otherwise expr_if_false
Description
IIf returns a different numeric or string or UDT value (not a reference) depending of the result of a conditional expression evaluated at run-time (evaluation at compile-time only if the comparison expression is a constant).
Its typical use is in the middle of an expression; it avoids splitting it to put a conditional in the middle.
IIf only evaluates the expression that it needs to return. This saves time, and can also be useful to prevent evaluating expressions that might be invalid depending on the
condition.
When IIf treats expressions of mixed numeric types (conditional expression evaluated at run-time):
- if at least one expression is of floating-point type, the result type is the floating-point type (the bigger in case of two floating-point types),
- if the two expressions are of integer types, the result type is the bigger type of both (see Coercion and Conversion for the precise ranking of integer types).
Example
Dim As Integer a, b, x, y, z
a = (x + y + IIf(b > 0, 4, 7)) \ z
is equivalent to:
Dim As Integer a, b, x, y, z, temp
If b > 0 Then temp = 4 Else temp = 7
a = (x + y + temp) \ z
Dim As Integer I
I = -10
Print I, IIf(I>0, "positive", IIf(I=0, "null", "negative"))
I = 0
Print I, IIf(I>0, "positive", IIf(I=0, "null", "negative"))
I = 10
Print I, IIf(I>0, "positive", IIf(I=0, "null", "negative"))
Sleep
Type UDT1
Dim As Integer I1
End Type
Type UDT2 Extends UDT1
Dim As Integer I2
End Type
Dim As UDT1 u1, u10 = (1)
Dim As UDT2 u2, u20 = (2, 3)
u1 = IIf(0, u10, u20)
Print u1.I1
u1 = IIf(1, u10, u20)
Print u1.I1
u2 = IIf(0 , u10, u20)
Print u2.I1; u2.I2
'u2 = Iif(1, u10, u20) ''Invalid assignment/conversion
Sleep
Dialect Differences
- Not available in the -lang qb dialect unless referenced with the alias __Iif.
Differences from QB
See also