Java 不能同时使用 Scanner.nextInt() 和 Scanner.nextLine()
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Can't use Scanner.nextInt() and Scanner.nextLine() together
提问by Umer Hassan
I have to get a string input and an integer input, but there order of input should be that integer comes first then user should be asked for string input
我必须得到一个字符串输入和一个整数输入，但输入的顺序应该是整数先出现然后用户应该被要求输入字符串
Scanner in = new Scanner(System.in);
input = in.nextLine();
k = in.nextInt();
in.close();
The above code works fine but if I take an integer input first like in the following code
上面的代码工作正常，但如果我像下面的代码一样首先输入一个整数
Scanner in = new Scanner(System.in);
k = in.nextInt();
input = in.nextLine();
in.close();
then it throws the java.lang.ArrayIndexOutOfBoundsException.
然后它抛出 java.lang.ArrayIndexOutOfBoundsException。
Here's the complete code of my source file:
这是我的源文件的完整代码：
import java.util.Scanner;
public class StringSwap {
公共类 StringSwap {
public static void main(String args[]) {
String input;
int k;
Scanner in = new Scanner(System.in);
k = in.nextInt();
input = in.nextLine();
in.close();
int noOfCh = noOfSwapCharacters(input);
originalString(input, noOfCh, k);
}
public static int noOfSwapCharacters(String s) {
char cS[] = s.toCharArray();
int i = 0, postCounter = 0;
while (cS[i] != '12345
hello
') {
if (cS[i] != '1 2 3 4 5 \n h e l l o \n
^
' && cS[i + 1] != '1 2 3 4 5 \n h e l l o \n
^
') {
cS[cS.length - 1 - postCounter] = '1 2 3 4 5 \n h e l l o \n
^
';
postCounter++;
}
i++;
}
return postCounter;
}
public static void originalString(String s, int noOfCh, int k) {
int counter = 1, chCounter = 0;
char cArray[] = s.toCharArray();
String post = "";
String pre = "";
String finalString = "";
char temp;
for (int i = 1; i <= k; i++) {
chCounter = 0;
counter = 1;
post = "";
pre = "";
for (int j = 0; j < cArray.length; j++) {
if (counter % 2 == 0 && chCounter <= noOfCh) {
temp = cArray[j];
post = temp + post;
cArray[j] = 'k = in.nextInt();
in.nextLine(); // Discard '\n'
input = in.nextLine();
';
chCounter++;
}
counter++;
}
for (int h = 0; h < cArray.length; h++) {
if (cArray[h] != '##代码##')
pre = pre + cArray[h];
}
finalString = pre + post;
for (int l = 0; l < finalString.length(); l++) {
cArray[l] = finalString.charAt(l);
}
}
System.out.println(finalString);
}
}
}
Kindly point out what I am doing wrong here.
请指出我在这里做错了什么。
采纳答案by dasblinkenlight
The problem is the '\n'character that follows your integer. When you call nextInt, the scanner reads the int, but it does not consume the '\n'character after it; nextLinedoes that. That is why you get an empty line instead of the string that you were expecting to get.
问题'\n'在于整数后面的字符。当您调用 时nextInt，扫描器会读取int，但不会消耗'\n'其后的字符；nextLine这样做。这就是为什么你得到一个空行而不是你期望得到的字符串。
Let's say your input has the following data:
假设您的输入具有以下数据：
##代码##Here is how the input buffer looks initially (^represents the position at which the Scannerreads the next piece of data):
这是输入缓冲区最初的样子（^表示Scanner读取下一条数据的位置）：
After nextInt, the buffer looks like this:
之后nextInt，缓冲区看起来像这样：
The first nextLineconsumes the \n, leaving your buffer like this:
第一个nextLine消耗\n，像这样离开你的缓冲区：
Now the nextLinecall will produce the expected result. Therefore, to fix your program, all you need is to add another call to nextLineafter nextInt, and discard its result:
现在nextLine调用将产生预期的结果。因此，要修复您的程序，您只需添加另一个对nextLineafter 的调用nextInt，并丢弃其结果：
