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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

By default bundled with:

IntelliJ IDEA 2024.1, Qodana for JVM 2024.1,

Can be installed with plugin:

Kotlin, 241.16690-IJ

Last modified: 29 April 2024