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下列程序的输出是什么？
class A { public static void main(String[] a) {
String v = “base”; v.concat(“ball”);
v.substring(1,5); System.out.println(v); }
}
分析：由于String是不可变的，当执行v.concat("ball")时，v本身还是指向“base”，当再执行v.substring(1,5)时，会报出界异常
如下为concat的源码：可以看到，最后返回的是新的字符串
public String concat(String str) { int otherLen = str.length(); if (otherLen == 0) { return this; } int len = value.length; char buf[] = Arrays.copyOf(value, len + otherLen); str.getChars(buf, len); return new String(buf, true); }
如下是substring源码：
public String substring(int beginIndex) { if (beginIndex < 0) { throw new StringIndexOutOfBoundsException(beginIndex); } int subLen = value.length - beginIndex; if (subLen < 0) { throw new StringIndexOutOfBoundsException(subLen); } return (beginIndex == 0) ? this : new String(value, beginIndex, subLen); } public String substring(int beginIndex, int endIndex) { if (beginIndex < 0) { throw new StringIndexOutOfBoundsException(beginIndex); } if (endIndex > value.length) { throw new StringIndexOutOfBoundsException(endIndex); } int subLen = endIndex - beginIndex; if (subLen < 0) { throw new StringIndexOutOfBoundsException(subLen); } return ((beginIndex == 0) && (endIndex == value.length)) ? this : new String(value, beginIndex, subLen); }
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原文地址：http://www.cnblogs.com/ljdblog/p/5952183.html
