问题
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
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
Scanner in = new Scanner(System.in);
k = in.nextInt();
input = in.nextLine();
in.close();
then it throws the java.lang.ArrayIndexOutOfBoundsException.
Here\'s the complete code of my source file:
import java.util.Scanner;
public class 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] != \'\\0\') {
if (cS[i] != \'\\0\' && cS[i + 1] != \'\\0\') {
cS[cS.length - 1 - postCounter] = \'\\0\';
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] = \'\\0\';
chCounter++;
}
counter++;
}
for (int h = 0; h < cArray.length; h++) {
if (cArray[h] != \'\\0\')
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.
回答1:
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; nextLine does that. That is why you get an empty line instead of the string that you were expecting to get.
Let's say your input has the following data:
12345
hello
Here is how the input buffer looks initially (^ represents the position at which the Scanner reads the next piece of data):
1 2 3 4 5 \n h e l l o \n
^
After nextInt, the buffer looks like this:
1 2 3 4 5 \n h e l l o \n
^
The first nextLine consumes the \n, leaving your buffer like this:
1 2 3 4 5 \n h e l l o \n
^
Now the nextLine call will produce the expected result. Therefore, to fix your program, all you need is to add another call to nextLine after nextInt, and discard its result:
k = in.nextInt();
in.nextLine(); // Discard '\n'
input = in.nextLine();
来源：https://stackoverflow.com/questions/23036062/cant-use-scanner-nextint-and-scanner-nextline-together