Java printing a String containing an integer

Question

I have a doubt which follows.

public static void main(String[] args) throws IOException{
  int number=1;
  System.out.println(\"M\"+number+1);
}

Answer 1

Try this:

System.out.printf("M%d%n", number+1);

Where %n is a newline

Answer 2

  System.out.println("M"+number+1);

String concatination in java works this way:

if the first operand is of type String and you use + operator, it concatinates the next operand and the result would be a String.

try

 System.out.println("M"+(number+1));

In this case as the () paranthesis have the highest precedence the things inside the brackets would be evaluated first. then the resulting int value would be concatenated with the String literal resultingin a string "M2"

Answer 3

System.out.println("M"+number+1);

Here You are using + as a concatanation Operator as Its in the println() method.

To use + to do sum, You need to Give it high Precedence which You can do with covering it with brackets as Shown Below:

System.out.println("M"+(number+1));

Answer 4

It has to do with the precedence order in which java concatenates the String,

Basically Java is saying

• "M"+number = "M1"
• "M1"+1 = "M11"

You can overload the precedence just like you do with maths

"M"+(number+1)

This now reads

• "M"+(number+1) = "M"+(1+1) = "M"+2 = "M2"

Answer 5

If you perform + operation after a string, it takes it as concatenation:

"d" + 1 + 1     // = d11 

Whereas if you do the vice versa + is taken as addition:

1 + 1 + "d"     // = 2d 

Answer 6

Try

System.out.println("M"+(number+1));