Scanner.nextLine（）返回null [英] Scanner.nextLine() return null

问题描述

我写过这段代码：

class PerfectPair {
    public static void main(String args[]) throws IOException{
        Scanner scan = new Scanner(System.in);
        int test=scan.nextInt();
        StringBuffer b=null;
        String a=null;
        while(test!=0){
            a=scan.nextLine();
            b=new StringBuffer(scan.nextLine());
            System.out.println(a);
            String reverse=b.reverse().toString();
            if((a.length()!=b.length()) || !(reverse.equals(a))){
                System.out.println("No");
            } 
            else 
            {   
                if((a.length()==b.length()) && (reverse.equals(a))) System.out.println("Yes");
            }

            --test;
        }
    }
}

输入的输入：

1
aa
ab

但变量a的值为null ..WHY ??请解释。也请更正代码，以便它读取完整的输入。

but the value of variable a is null ..WHY?? Please explain .Also please correct the code so that it reads full input.

推荐答案

那是因为你输入1然后输入。因此，您的nextLine方法调用只读取返回键，而nextInt只读取整数值而忽略返回键。要避免此问题：

That's because you are entered 1 followed by an enter . So your nextLine method call just reads return key while nextInt just reads integer value ignoring the return key. To avoid this issue:

在阅读输入后，您会调用以下内容：

Just after reading input, you call something like:

int test=scan.nextInt();
scan.nextLine();//to read the return key.

如果你想避免这种情况，那我建议你读完整行然后将其转换为整数。有些事情如下：

If you want to avoid that as well, then i would suggest, you read the whole line and then convert it to integer. Some thing like:

int test=Integer.parseInt(scan.nexLine());

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