Sc.nextLine() and sc.nextInt() in idea deal with the problem of carriage return and line break

Article directory

1. Find the problem
- 1. sc. nextLine()
- 2. sc. nextInt()
2. Analyzing the problem
3. Summary
4. New issues
- 1.
- 2.

1. Found the problem

Today, when I checked the questions in Luogu, I found that the string could not be entered, and then I came up with today’s article. Look at the following question first.

1. sc. nextLine()

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);

        String str1 = sc. nextLine();
        String str2 = sc. nextLine();

        System.out.println(str1);
        System.out.println(str2);

    }

}

Pretty basic, right, but the results are amazing, let’s take a look at the results of the execution.

Here I entered the strings “1” and “2”. When printing, only “1” is printed. Why, let’s debug it. I saw the following results:

str2 did not read the “2” it should have. Guess: If it is not read into a different character? Let’s modify the code as follows:

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);

        String str1 = sc. nextLine();
        sc. nextLine();
        String str2 = sc. nextLine();

        System.out.println(str1);
        System.out.println(str2);

    }

}

The blogger added a sc.nextLine() here to read other characters

This time str2 holds the correct value and prints it out successfully.

2. sc. nextInt()

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);

        int num = sc. nextInt();

        String str = sc. nextLine();

        System.out.println(num);
        System.out.println(str);

    }

}

I feel that there is nothing wrong with it. It turned out to be running. Is it this? As shown below:

Just let me enter a num, and before I had time to enter it, it was over. Let’s debug it, as shown below:

As in the above case, str is still not entered, and there is one null value. That’s the same as above, we also add a sc.nextLine() to see how the result is:

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);

        int num = sc. nextInt();
        sc. nextLine();
        
        String str = sc. nextLine();

        System.out.println(num);
        System.out.println(str);

    }

}

The result of the operation is as follows:

This time I was asked to enter str, but it seems that it is still not saved. Why is this a bit like the situation in sc, nextLine() above. . . debug again:

Still haven’t read it in, well, let’s add another sc.nextLine() to try, the code is as follows:

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);

        int num = sc. nextInt();
        sc. nextLine();
        sc. nextLine();
        
        String str = sc. nextLine();

        System.out.println(num);
        System.out.println(str);

    }

}

This time the operation was successful, as shown below:

2. Analyzing the problem

Why does this happen? I checked the information and found:

After pressing the Enter key on the keyboard, two characters are actually sent to the program ———‘\r’ and ‘\
‘

sc.nextInt(), after entering the number, both ‘\r’ and ‘\
‘ are left in the buffer.
- After sc.nextInt(), two nextLine() are needed to handle other characters.
And sc.nextLine(), after entering the string, only ‘\
‘ remains in the buffer.
- After sc.nextLine(), a nextLine() is needed to handle other characters.

Three. Summary

In the Unix system, the end of each line is only ““, that is, “\
“; In the Windows system, the end of each line is ““, that is, “\r\ n”; in the Mac system, the end of each line is ““, that is, “\r”
‘\r’means carriage return (carriage return), that is, returning to the beginning of the line, and does not include the action of line break;’\
‘means line feed (line feed), that is, moves to a new line (next line)
nextInt(), next(), nextFloat(), and nextDouble() all only read valid characters, and will not read space bar, tab key and enter key.
- It will auto-skip when it encounters these invalid characters before entering valid characters, and after valid characters End reading, leaving invalid characters in buffer.
nextLine() can read everything except the carriage return, that is, the reading ends when encountering the carriage return, but the carriage return will be left in the buffer.

refer to:
https://blog.csdn.net/cnds123/article/details/126389380
https://blog.csdn.net/weixin_55000908/article/details/123888389
https://blog.csdn.net/m0_63162560/article/details/126211023

Four. New questions

1.

package Luogu;

import java.util.*;

public class Test {<!-- -->

    public static void main(String[] args){<!-- -->
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter the content you want to print:");
        String info1 = sc. nextLine();
        System.out.println("The digital data entered by the keyboard is: " + info1);
        String info2 = sc. nextLine();
        String info3 = sc. nextLine();
        System.out.println("The string data entered by the keyboard is: " + info2);
        System.out.println("The digital data entered by the keyboard is: " + info3);


    }

}

The result is as follows:

Here info2 actually read in? ? ? ? Stumped System.out.println (); has the effect of clearing the temporary storage area? The blogger read the source code for a long time, but didn’t understand it. . . Maybe it’s too good

2.

There is also such a piece of code, where sc.nextLine() is added, but it will wait for input. . It’s puzzling.

If anyone knows, please feel free to enlighten me, clap your fists!