Explain println and print statement after a loop has been executed in Java?

Question

What happens here, that is when print is used, why will it not print where the line stops?

for(int i = 0; i <=2; i++){
System.out.println(i)
System.out.print("s");
}

Why will it not print the s after 2 like this:

0

1

2s

From what i learnt, someone said it will buffer for ever? What does that mean? The computer will know to print the letter s beside 2 because it has stopped there, so why not print?

Answer 1

You are missing a semi-colon after the first statement in your loop.

for (int i = 0; i <=2; i++) {
    System.out.println(i);
    System.out.print("s");
}

The above code will output this:

0
s1
s2
s

However if you do this:

for (int i = 0; i <=2; i++) {
    System.out.print(i);
    System.out.println("s");
}

Which will print the following:

0s
1s
2s

With an extra line break at the end.

Answer 2

For each iteration in the loop, the code prints the integer, adds a newline (since println is used) and prints the s. So you will have an output like

0

s1

s2

s

Answer 3

What happens with you code:

0
s1 
s2
s

Why it happens: if you use

print("s");

the program will print "s" and thats it. if you use:

println("s");

you'll get a new line after each "s" which I suppose is what you seek.

Answer 4

这是因为println将在语句的末尾而不是语句的开头保留换行符（\\ n）。

Answer 5

Actually print prints the s where the line stops!. The problem here is, println already put a new line first, and then print starts from that place.

Answer 6

Both statements are in the same loop so will run one after another.

So:

System.out.println(i);   

will give i(whatever that variable is) then a new line.

System.out.print("s");   

will print "s". With no new line.

Your output will be

0
s1
s2
s

And since your only want an output on the last iteration of the loop, you need an 'if':

    for (int i = 0; i <= 2; i++) {
        System.out.print(i);
        if (i == 2) {
        System.out.print("s");
        } else {
        System.out.println("");
        }
    }

Which gives:

0
1
2s

Answer 7

The actual answer is that the previous println already puts it to the next line. Using System.out.print("s") actually goes horizontally forever, therefore it never prints.