我想知道 system.out.println() 在这段代码中做了什么

Question

I don't know why there are system.out.println in the end of these code. what is this and why is it here?

This code is a code from my friend, he told me to understand this code and I don't understand why there are system.out.println after this code.

public class Nestedlooplab2 {
    public static void main(String[] args) {
        for (byte i = 1; i <= 5; i++) {
            for (byte j = 1; j <= i; j++) {
                System.out.print(".");
            }
            for (byte k=1; k<=(5-i);k++) {
                System.out.print("*");
            } System.out.println();}
    }
    }

Answer 1

Welcome to stack overflow!

By itself a System.out.println() will just print a newline character, eg \n . If you were to add a parameter to this statement it would print your parameter along with a newline.

Here are the JavaDocs for println

Terminates the current line by writing the line separator string. The line separator string is defined by the system property line.separator, and is not necessarily a single newline character ('\n').

The code you linked would print a certain amount of . characters and certain amount of * characters without any line breaks. The final System.out.println(); would then print a newline character and the loop would start over again.

Output:

.****
..***
...**
....*
.....

Answer 2

The output of your code is:

.****
..***
...**
....*
.....

Without the System.out.println(); , the output would be:

.****..***...**....*.....

The System.out.println(); prints a new line character to the screen for each interaction of the loop.

Answer 3

First, let's take a look at the output of your code:

.****
..***
...**
....*
.....

As a String, this is the same as:

.****\n..***\n...**\n....*\n.....\n

Using System.out in Java writes to the standard output-stream of characters by default. When you write to this stream, you can use various methods, including System.out.print and System.out.println . Calling print will output the exact String that you provide to it, whereas calling println will output the String you provide to it, followed by a new line (the line separator for your system). If you call System.out.println() ( println with no parameter), you will output the String you provided ( "" ) as well as move the output to the next line. Essentially, this means removing calls to System.out.println() in your code will result in the following output:

.****..***...**....*.....

This output will look exactly the same as a String. As you can see, there are no newline characters ( \n ) within your output when you only call System.out.print and don't call System.out.println .

Finally, let's take a look at your code in the context of making it easier to read and understand. I'm using the Java 11+ feature String.repeat to massively simplify the operation of repeating a String here:

public static void main(String[] args) {
  for (int i = 1; i <= 5; i++) {
    System.out.println(".".repeat(i)+"*".repeat(5-i));
  }
}

Output:

.****
..***
...**
....*
.....

This is equivalent to the original output. However, it's much clearer to read and understand what is going on. Supposing that you don't have access to those features, you can do the following instead:

for (int i = 1; i <= 5; i++) {
  for(int rpts = 0; rpts < i; rpts++) {
    System.out.print('.');
  }
  for(int rpts = i; rpts < 5; rpts++) {
    System.out.print('*');
  }
  System.out.println();
}

This snippet of code also has the same output. Its content is not much different from your code snippet, as your code snippet does have the right idea. However, it is more consistently formatted, which makes it easier for yourself and others to read the code. Do note how the repeats are labelled with a name rpts , and (in both examples) the iteration variables are ints. Java programmers usually use ints to iterate because the space you save from using bytes to iterate is negligible enough to ignore for most applications, and integers cover most ranges of values you might want to iterate over.