java – 在catch中返回的语句的行为,最后
发布时间:2020-12-14 05:24:24 所属栏目:Java 来源:网络整理
导读:public class J { public Integer method(Integer x) { Integer val = x; try { return val; } finally { val = x + x; } } public static void main(String[] args) { J littleFuzzy = new J(); System.out.println(littleFuzzy.method(new Integer(10)));
public class J { public Integer method(Integer x) { Integer val = x; try { return val; } finally { val = x + x; } } public static void main(String[] args) { J littleFuzzy = new J(); System.out.println(littleFuzzy.method(new Integer(10))); } } 它将返回“10”. 现在我只是将Return类型的整数替换为StringBuilder,而Output被更改. public class I { public StringBuilder method(StringBuilder x) { StringBuilder val = x; try { return val; } finally { val = x.append("aaa"); } } public static void main(String[] args) { I littleFuzzy = new I(); System.out.println(littleFuzzy.method(new StringBuilder("abc"))); } } OutPut是“abcaaa” 所以,任何人都可以详细解释一下. 解决方法
只是因为整数不可变,所以后方法返回即使值在方法中更改它也不反映,并且反映在StringBuilder对象
编辑: public class J { public String method(String x) { String val = x; try { return val; } finally { val = x + x; } } public static void main(String[] args) { J littleFuzzy = new J(); System.out.println(littleFuzzy.method("abc")); } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |