IIf
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. 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

- New to FreeBASIC

See also

- `If...Then`