在Java中连接空字符串

参见英文答案 > String concatenation with Null                                    3个
为什么以下工作？我希望抛出NullPointerException.

String s = null;
s = s + "hello";
System.out.println(s); // prints "nullhello"

解决方法:

为什么一定有用？

导致JLS 8, § 5.1.11 “String Conversion”的JLS 5, Section 15.18.1.1 JLS 8 § 15.18.1 “String Concatenation Operator +”要求此操作成功而不会失败：

…Now only reference values need to be considered. If the reference is null, it is converted to the string “null” (four ASCII characters n, u, l, l). Otherwise, the conversion is performed as if by an invocation of the toString method of the referenced object with no arguments; but if the result of invoking the toString method is null, then the string “null” is used instead.

它是如何工作的？

我们来看看字节码吧！编译器接受你的代码：

String s = null;
s = s + "hello";
System.out.println(s); // prints "nullhello"

并将其编译为字节码,就像你写了这样：

String s = null;
s = new StringBuilder(String.valueOf(s)).append("hello").toString();
System.out.println(s); // prints "nullhello"

(您可以使用javap -c自己完成)

StringBuilder的append方法都处理null就好了.在这种情况下,因为null是第一个参数,所以调用String.valueOf(),因为StringBuilder没有采用任意引用类型的构造函数.

如果您已经完成了s =“hello”,则等效代码将为：

s = new StringBuilder("hello").append(s).toString();

在这种情况下,append方法获取null,然后将其委托给String.valueOf().

注意：字符串连接实际上是编译器决定执行哪些优化的罕见地方之一.因此,“完全等效”代码可能因编译器而异. JLS, Section 15.18.1.2允许此优化：

To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.

我用来确定上面“等效代码”的编译器是Eclipse的编译器,ecj.

标签：java,string,concatenation,string-concatenation
来源： https://codeday.me/bug/20190911/1804638.html