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
Available in:	IntelliJ IDEA 2023.3, Qodana for JVM 2023.3
Plugin:	Kotlin, @snapshot@

Last modified: 13 July 2023