Java中++i与i++的效率对比：是否与C++情况一致？

Great question! The short answer is: for most practical purposes, no—there's barely any performance difference between ++i and i++ in Java, and modern JVMs often optimize them to identical code. Let's break down why this differs from C++:

1. For primitive types (int, long, etc.) #

In C++, the performance gap between ++i and i++ becomes noticeable with user-defined types (like iterators), where i++ has to create a copy of the current value to return before incrementing. But in Java:

• Primitive types are not objects, so there's no "copy" overhead to worry about.
• i++ works by reading the current value, incrementing the variable, then returning the original value. ++i increments first, then returns the updated value.
• Modern JIT compilers (like HotSpot) will optimize these operations to nearly identical machine code, especially in loops (the most common place you'll use these operators). For example, a for loop using i++ vs ++i will run at the exact same speed.

2. For wrapper types (Integer, Long, etc.) #

Wrapper types add a layer of boxing/unboxing, but even here, the difference is negligible:

• Both i++ and ++i for wrappers are compiled to code that unboxes the primitive value, increments it, then reboxes it (using Integer.valueOf() or similar).
• The steps involved are identical in terms of overhead—there's no extra copy created for i++ like there would be in C++ for custom types. The only potential variation is if the reboxed value falls into the wrapper's cache range (e.g., -128 to 127 for Integer), but that's unrelated to the increment operator itself.

Key takeaway #

Unless you're working in an extremely performance-sensitive scenario (think billions of iterations), you'll never notice a difference between ++i and i++ in Java. Focus on code readability instead—use whichever makes your intent clearer (e.g., i++ if you need the original value, ++i if you only care about the updated value).

内容的提问来源于stack exchange，提问作者Raul Luna