我刚刚创建了一个程序,它应该扫描文件并计算行数,元音数和其他许多东西,但是当我运行它时,它会进入一个无限循环。我很确定它与我不熟悉的.hasNext()
和.hasNextLine()
方法有关
import java.util.*;
import java.io.*;
/**
*
*
*/
public class Wordcount1 {
public static void main(String[] args) {
int vowels=0;
int punctuation=0;
int sentences=0;
int words=0;
int lines=0;
int alphaNumeric=0;
try{
Scanner input = new Scanner(System.in);
System.out.println("Enter file name: ");
File file = new File(input.nextLine());
Scanner fileReader = new Scanner(file);
while(fileReader.hasNextLine()){
lines +=1;
}
while(fileReader.hasNext()){
{ fileReader.next();
words +=1;
}
String word = fileReader.next();
for (int i=0; i<word.length();i++){
char ch= word.charAt(i);
if(ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u')
vowels +=1;
if((ch=='!'||ch=='.'||ch=='?'))
sentences +=1;
if(Character.isLetterOrDigit(ch))
alphaNumeric +=1;
switch(ch){
case ',':
punctuation +=1;
break;
case '[':
punctuation +=1;
break;
case ']':
punctuation +=1;
break;
case ':':
punctuation +=1;
break;
case '`':
punctuation +=1;
break;
case '-':
punctuation +=1;
break;
case '!':
punctuation +=1;
break;
case '_':
punctuation +=1;
break;
case '(':
punctuation +=1;
break;
case ')':
punctuation +=1;
break;
case '.':
punctuation +=1;
break;
case '?':
punctuation +=1;
break;
case '"':
punctuation +=1;
break;
case ';':
punctuation +=1;
break;
}
}
}
System.out.println("The number of words in the file name: " + words);
System.out.println("The number of lines in the file name: " + lines);
System.out.println("The number of alphanumeric characters: "
+ "in the file name: " + alphaNumeric);
System.out.println("The number of sentences in the file name: "
+ sentences);
System.out.println("The number of vowels in the file name: " + vowels);
System.out.println("The number of punctuations in the file name: "
+ punctuation);
}catch(FileNotFoundException e){
e.printStackTrace();
}
}
}
答案 0 :(得分:2)
while(fileReader.hasNextLine()){
lines +=1;
}
在这里,你反复询问文件“你还有更多行”,但实际上并没有消耗任何行。所以,它一直回答“是的,我有更多的行”。
答案 1 :(得分:0)
因为您没有在fileReader上调用next()
方法
调用Scanner.hasNext()仅表示是否有令牌,但不会使扫描程序前进: