Java Scanner输入异常：获取数组长度后跳过首个字符串输入

这是Java新手使用Scanner时经常踩的小坑，我来帮你捋清楚问题根源和解决办法～

问题重现 #

在以下Java代码中，首次循环时无法获取第一个单词的输入。当不向用户获取字符串数组长度时，代码运行正常。可查看输出，其中未获取word 0的输入。

用户提供的代码：

public class Stringinput { 
    public static void main(String[] args) { 
        Scanner in = new Scanner(System.in); 
        System.out.println("How many words do you need to make the sentence: "); 
        int n = in.nextInt(); 
        String[] names = new String[n]; 
        for (int i = 0; i < names.length; i++) { 
            System.out.printf("Enter word no %d : ",i); 
            names[i]= in.nextLine(); 
        } 
        System.out.println("The sentence you made is: "); 
        for (int i = 0; i < names.length; i++) { 
            System.out.print(names[i]+" "); 
        } 
        System.out.println(""); 
    } 
}

运行输出：

How many words do you need to make the sentence: 4 
Enter word no 0 : 
Enter word no 1 : My 
Enter word no 2 : name 
Enter word no 3 : is 
The sentence you made is:  My name is  
BUILD SUCCESSFUL (total time: 11 seconds)

问题根源 #

核心原因是**nextInt()不会读取输入末尾的换行符**：
当你输入4然后按下回车时，in.nextInt()只取走了数字4，而回车产生的\n字符还留在输入缓冲区里。紧接着第一次循环调用in.nextLine()时，它会直接读取这个残留的换行符，把它当成一个空字符串赋值给names[0]，所以看起来像是跳过了第一个单词的输入。

两种解决方案 #

方案1：在nextInt()后额外调用一次nextLine()清空换行符

在读取完整数n之后，加一行in.nextLine()来吃掉残留的换行符，这样后续的nextLine()就能正常读取用户输入的单词了：

public class Stringinput { 
    public static void main(String[] args) { 
        Scanner in = new Scanner(System.in); 
        System.out.println("How many words do you need to make the sentence: "); 
        int n = in.nextInt(); 
        // 新增这一行，清空输入缓冲区里的换行符
        in.nextLine();
        String[] names = new String[n]; 
        for (int i = 0; i < names.length; i++) { 
            System.out.printf("Enter word no %d : ",i); 
            names[i]= in.nextLine(); 
        } 
        System.out.println("The sentence you made is: "); 
        for (int i = 0; i < names.length; i++) { 
            System.out.print(names[i]+" "); 
        } 
        System.out.println(""); 
    } 
}

方案2：用nextLine()读取整行后转成整数（更稳妥）

避免混用nextInt()和nextLine()，直接用nextLine()读取用户输入的整行内容，再通过Integer.parseInt()转成整数，这样就不会有换行符残留的问题：

public class Stringinput { 
    public static void main(String[] args) { 
        Scanner in = new Scanner(System.in); 
        System.out.println("How many words do you need to make the sentence: "); 
        // 读取整行后转成整数
        int n = Integer.parseInt(in.nextLine()); 
        String[] names = new String[n]; 
        for (int i = 0; i < names.length; i++) { 
            System.out.printf("Enter word no %d : ",i); 
            names[i]= in.nextLine(); 
        } 
        System.out.println("The sentence you made is: "); 
        for (int i = 0; i < names.length; i++) { 
            System.out.print(names[i]+" "); 
        } 
        System.out.println(""); 
    } 
}

修改后运行效果 #

调整后再运行，输入4个单词的话，输出会正常获取所有输入：

How many words do you need to make the sentence: 4
Enter word no 0 : Hello
Enter word no 1 : My
Enter word no 2 : name
Enter word no 3 : is
The sentence you made is: Hello My name is 

内容的提问来源于stack exchange，提问作者 Mahbub