1. 程式人生 > >java.lang.NumberFormatException: For input string: "19000000000" at java.lang.NumberFormatExceptio

java.lang.NumberFormatException: For input string: "19000000000" at java.lang.NumberFormatExceptio

Exception in thread "main" java.lang.NumberFormatException: For input string: "19000000000"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at num1.main(num1.java:13)

1.原始碼如下

String s="19000000000";

    //   String m=s.substring(0, s.length());
       System.out.println(s);
       int f=Integer.parseInt(s);

       System.out.println(f);

將String s="19000000000";改成String s="1900000000"就不會報錯了,原因是因為用Interger.parseInt()解析的數不能太大