我得到了这个例外:
x :: IO String
x = return "hello world"
main = do
x >>= print
使用此代码时:
Exception in thread "main" java.lang.NumberFormatException: For input string: "55 45 65 88 "
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.valueOf(Unknown Source)
示例输入:
4
55 45 65 88(在这里,当我按下输入时,它给出了上述错误)
答案 0 :(得分:4)
StringTokenizer
不支持正则表达式。
StringTokenizer tokens = new StringTokenizer(line, "\\s+");
// This will look for literal "\s+" string as the token.
请改用
StringTokenizer tokens = new StringTokenizer(line, " "); // Just a space.
编辑:正如@MasterOdin所指出的,StringTokenizer
的默认分隔符是空格" "
。因此,下面也会以同样的方式工作,
StringTokenizer tokens = new StringTokenizer(line);
答案 1 :(得分:2)
你可以采用简单的方式:
String []m=br.readLine().split(" "); // split the line delimited with space as array of string.
for(int i=0;i<m.length;i++){
marks.add(Integer.valueOf(m[i])); // add to the marks array list
}
编辑:根据T.G
for (String s : br.readLine().split("\\s+")) {
marks.add(Integer.valueOf(s));
}