String concat和Java中的+运算符有区别吗？ [英] Is there a difference between String concat and the + operator in Java?

问题描述

java字符串连接

我很好奇两者之间有什么区别。

I'm curious what is the difference between the two.

我理解字符串池的方式如下：

The way I understand the string pool is this:

这会在字符串池中创建3个字符串对象，其中2个字符串对象丢失。

This creates 3 string objects in the string pool, for 2 of those all references are lost.

String mystr = "str";
mystr += "end";

这不会在字符串池中创建3个对象吗？

Doesn't this also create 3 objects in the string pool?

String mystr = "str";
mystr = mystr.concat("end")

我知道StringBuilder和StringBuffer很多当存在大量的连接时，在内存使用方面更有效。我只是好奇，如果在内存使用方面+运算符和concat之间有任何区别。

I know StringBuilder and StringBuffer are much more efficient in terms of memory usage when there's lots of concatination to be done. I'm just curious if there's any difference between the + operator and concat in terms of memory usage.

推荐答案

没有区别这个特例; 但是，它们通常不一样。

There's no difference in this particular case; however, they're not the same in general.

str1 + = str2 相当于执行以下操作：

str1 += str2 is equivalent to doing the following:

str1 = new StringBuilder().append(str1).append(str2).toString();

为了向自己证明这一点，只需制作一个带两个字符串且 + = 是第二个字符串，然后检查反汇编的字节码。

To prove this to yourself, just make a simple method that takes two strings and +='s the first string to the second, then examine the disassembled bytecode.

相比之下， str1 .concat（str2）只需创建一个新字符串，它是 str1 和 str2 的串联，对于少量串联字符串来说成本较低（但是对于第一种方法会丢失较大的数字）。

By contrast, str1.concat(str2) simply makes a new string that's the concatenation of str1 and str2, which is less expensive for a small number of concatenated strings (but will lose to the first approach with a larger number).

此外，如果 str1 为空，请注意 str1.concat（str2）抛出一个NPE，但 str1 + = str2 只会将 str1 视为无效，而不会抛出异常。 （也就是说，它产生null与 str2 的值连接。如果 str2 是，那么，foo ，你最终会得到nullfoo。）

Additionally, if str1 is null, notice that str1.concat(str2) throws a NPE, but str1 += str2 will simply treat str1 as if it were null without throwing an exception. (That is, it yields "null" concatenated with the value of str2. If str2 were, say, "foo", you would wind up with "nullfoo".)

更新：参见 此StackOverflow问题 ，几乎相同。

Update: See this StackOverflow question, which is almost identical.

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