If-Then foldable to '?.'
Reports if-then
expressions that can be folded into safe-access (?.
) expressions.
Example:
fun bar(x: String) = ""
fun foo(a: String?) {
if (a != null) bar(a) else null
}
The quick fix converts the if-then
expression into a safe-access (?.
) expression:
fun bar(x: String) = ""
fun foo(a: String?) {
a?.let { bar(it) }
}
Inspection Details | |
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Available in: | IntelliJ IDEA 2023.3, Qodana for JVM 2023.3 |
Plugin: | Kotlin, @snapshot@ |
Last modified: 13 July 2023