Kotlin技术疑问：使用val替代var是否存在运行时性能差异？

Great question—since you already get the semantic difference between var (mutable) and val (read-only), let’s cut straight to the runtime performance angle you’re focused on.

Short Answer #

There’s no meaningful runtime performance gap between var and val when the variable is never reassigned. Swapping var for val won’t make your code run faster in any noticeable way.

Longer Breakdown #

Here’s why the performance impact is negligible:

• Bytecode Compilation: For local variables, Kotlin compiles unused-reassigned var and val to nearly identical Java bytecode. The compiler might mark val as final in the bytecode, but this is a compile-time hint, not a runtime constraint that directly boosts speed.
• JVM JIT Optimization: The JVM’s Just-In-Time compiler is smart enough to detect when a var is never actually modified during execution. It will optimize that variable exactly like a val—treating it as effectively immutable, unlocking the same low-level optimizations (like constant folding, inlining, etc.).
• Class Properties: For class-level properties, val generates a private final field with only a getter, while var adds a setter. But if you never call the setter for a var, the runtime overhead is zero—the JIT will eliminate any unused code paths entirely.

The Real Win with val #

The value of val isn’t performance—it’s compile-time safety and code clarity. By using val, you’re telling the compiler (and other developers) that this variable shouldn’t be changed, which prevents accidental reassignment and makes your code easier to reason about. Any "performance gain" you might hear about is a side effect of this compile-time guard, not a direct runtime improvement.

Quick Example #

Take these two functions:

fun useVar() {
    var greeting = "Hi there"
    // greeting is never reassigned
    println(greeting)
}

fun useVal() {
    val greeting = "Hi there"
    println(greeting)
}

When compiled and optimized by the JVM, these will run at exactly the same speed. The JIT doesn’t care about your declaration choice—it only cares if the value is actually modified.

内容的提问来源于stack exchange，提问作者J K