还是老问题：String创建对象的数量

踏雁寻花 2014-03-21 01:31:32

String result = foo() + arg;

result += boo();

System.out.println(“result = “ + result);

创建了几个StringBuilder对象。
引用网上的一个回复：
In this example we have 3 StringBuilders allocated in the background – one for each plus operation, and two additional Strings – one to hold the result of the second assignment and another to hold the string passed into the print method. That’s 5 additional objects in what would otherwise appear to be a pretty trivial statement.
这个回答是否有道理。

另外：

StringBuilder value = new StringBuilder(“result = “);

value.append(foo()).append(arg).append(boo());

System.out.println(value);

又会有几个StringBuilder对象？
还是引用：
allocating only one StringBuilder and one String to hold the final result, instead of the original five objects.
看下是否有道理？

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lxbccsu 2014-03-24

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引用 7 楼 tayanxunhua 的回复:

[quote=引用 4 楼 lxbccsu 的回复:] [quote=引用楼主 tayanxunhua 的回复:]
String result = foo() + arg;
result += boo();
System.out.println(“result = “ + result);
创建了几个StringBuilder对象。引用网上的一个回复： In this example we have 3 StringBuilders allocated in the background – one for each plus operation, and two additional Strings – one to hold the result of the second assignment and another to hold the string passed into the print method. That’s 5 additional objects in what would otherwise appear to be a pretty trivial statement. 这个回答是否有道理。另外：
StringBuilder value = new StringBuilder(“result = “);
value.append(foo()).append(arg).append(boo());
System.out.println(value);
又会有几个StringBuilder对象？还是引用： allocating only one StringBuilder and one String to hold the final result, instead of the original five objects. 看下是否有道理？

上面说的都是对的；首先Java中是不可以重载操作符，但这个 + 能用于字符串操作，也就是重载了操作符，这也是Java中唯一操作符重载的地方；其次，任何字符串的 + 操作都会转换为StringBuilder的操作；但你若这样操作： StringBuilder value = new StringBuilder(“result = “ + foo()); value.append(arg + boo()); 那你又是白优化了，还是会创建3个 StringBuilder对象； [/quote] "allocating only one StringBuilder and one String to hold the final result,"这里的"one String to hold the final result"指的是哪个？[/quote] 这里说的是StringBuilder 的toString方法，这个方法会把StringBuilder的内容转变为String，当你调用System.out.println(value)时，就会把这个最终的结果打印出来；

king_1993 2014-03-24

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好深奥的样子

代码间的舞者 2014-03-22

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学习了，标记下

踏雁寻花 2014-03-21

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引用 4 楼 lxbccsu 的回复:

[quote=引用楼主 tayanxunhua 的回复:]
String result = foo() + arg;
result += boo();
System.out.println(“result = “ + result);
创建了几个StringBuilder对象。引用网上的一个回复： In this example we have 3 StringBuilders allocated in the background – one for each plus operation, and two additional Strings – one to hold the result of the second assignment and another to hold the string passed into the print method. That’s 5 additional objects in what would otherwise appear to be a pretty trivial statement. 这个回答是否有道理。另外：
StringBuilder value = new StringBuilder(“result = “);
value.append(foo()).append(arg).append(boo());
System.out.println(value);
又会有几个StringBuilder对象？还是引用： allocating only one StringBuilder and one String to hold the final result, instead of the original five objects. 看下是否有道理？

zhjdg 2014-03-21

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我对 + += 不了解，所以第一个不知道。第2个对 one String to hold the final result，我不同意。 System.out.println(value); value是StringBuilder 看源代码一清二楚。

-江沐风- 2014-03-21

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http://bbs.csdn.net/topics/270015407

lxbccsu 2014-03-21

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引用楼主 tayanxunhua 的回复:

String result = foo() + arg;
result += boo();
System.out.println(“result = “ + result);
创建了几个StringBuilder对象。引用网上的一个回复： In this example we have 3 StringBuilders allocated in the background – one for each plus operation, and two additional Strings – one to hold the result of the second assignment and another to hold the string passed into the print method. That’s 5 additional objects in what would otherwise appear to be a pretty trivial statement. 这个回答是否有道理。另外：
StringBuilder value = new StringBuilder(“result = “);
value.append(foo()).append(arg).append(boo());
System.out.println(value);
又会有几个StringBuilder对象？还是引用： allocating only one StringBuilder and one String to hold the final result, instead of the original five objects. 看下是否有道理？

-江沐风- 2014-03-21

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StringBuffer和StringBuilder基本上差不多，就是线程的问题； String定义的是常量，即String实例化以后所有的属性都是final的，而StringBuffer和StringBuilder确不是，这就是所谓的可变与不可变；一个比喻：String是一个商品； StringBuffer/StringBuilder是生产这个商品的流水线； StringBuffer速度慢，但（线程）安全性高； StringBuilder速度快，但（线程）安全性差；