为什么在 Java 中使用 StringBuffer 而不是字符串连接运算符 [英] Why to use StringBuffer in Java instead of the string concatenation operator

问题描述

有人告诉我在 Java 中使用 StringBuffer 来连接字符串比使用 + 运算符来连接 String 的效率更高.当你这样做时，引擎盖下会发生什么?StringBuffer 有何不同?

Someone told me it's more efficient to use StringBuffer to concatenate strings in Java than to use the + operator for Strings. What happens under the hood when you do that? What does StringBuffer do differently?

推荐答案

现在，在几乎所有情况下，最好使用 StringBuilder(它是一个非同步版本；什么时候并行构建字符串?)，但会发生以下情况:

It's better to use StringBuilder (it's an unsynchronized version; when do you build strings in parallel?) these days, in almost every case, but here's what happens:

当你对两个字符串使用 + 时，它会编译这样的代码:

When you use + with two strings, it compiles code like this:

String third = first + second;

对于这样的事情:

StringBuilder builder = new StringBuilder( first );
builder.append( second );
third = builder.toString();

因此，对于一些小例子，它通常没有什么区别.但是当你构建一个复杂的字符串时，你通常需要处理的远不止这些；例如，您可能会使用许多不同的附加语句，或者像这样的循环:

Therefore for just little examples, it usually doesn't make a difference. But when you're building a complex string, you've often got a lot more to deal with than this; for example, you might be using many different appending statements, or a loop like this:

for( String str : strings ) {
  out += str;
}

在这种情况下，一个新的 StringBuilder 实例和一个新的 String(out 的新值 - Strings 是不可变的)在每次迭代中都是必需的.这是非常浪费的.用一个 StringBuilder 代替它意味着你可以只生成一个 String 而不用你不关心的 String 填充堆.

In this case, a new StringBuilder instance, and a new String (the new value of out - Strings are immutable) is required in each iteration. This is very wasteful. Replacing this with a single StringBuilder means you can just produce a single String and not fill up the heap with Strings you don't care about.

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