System.out.println behavior when using String and int together

Question

Consider the below code snippet -

public class Student {

    public static void main(String args[]) {
        int a = 3;
        int b = 4;
        System.out.println(a + b +" ");
        System.out.println(a + b +" "+ a+b);
        System.out.println(""+ a+b);

    }
}

The output for the above snippet is coming as -

7
7 34
34

It is clear from the output that if we use String at first in the print statement then the integers are concatenated. But, if we use integer at first then the values are added and displayed.

Can someone please explain why is this behavior?

I even tried to look at the implementation of println() method in PrintStream class but could not figure out.

what you have described, that IS the behavior. the concatenation will only know it's a String concatenation once it detects a String. do the adding of int's before a String, and it'll execute that augmentation before it passes it's first String — Stultuske, Mar 18 '19 at 07:41
This has nothing to do with `println`, instead, this is how expressions are resolved in Java. — MC Emperor, Mar 18 '19 at 07:50

cнŝdk · Accepted Answer · 2019-03-18T14:36:39.567

Actually it's not println() implementation who's causing that, this is Java way to treat the + operator when dealing with Strings.

In fact the operation is treated from left to right so:

If the string comes before int (or any other type) in the operation String conversion is used and all the rest of the operands will be treated as strings, and it consists only of a String concatenation operation.
If int comes first in the operation it will be treated as int, thus addition operation is used.

That's why a + b +" " gives 7 because Stringis in the end of the operation, and for other expressions a + b +" "+ a+b or ""+ a+b, the variables a and b will be treated as strings if the come after a String in the expression.

For further details you can check String Concatenation Operator + in Java Specs.

Eran · Answer 2 · 2019-03-18T07:49:29.813

It has nothing to do with the implementation of println. The value of the expression passed to println is evaluated before println is executed.

It is evaluated left to right. As long as both operands of + are numeric, addition will be performed. Once at least one of the operands is a String, String concatenation will be performed.

System.out.println(a    +      b          +       " ");
                  // int + int      int + String
                  // addition,      concatenation 
System.out.println(a    +      b    +    " "   +    a   +    b);
                  // int + int  int + String Str+Str  Str+Str
                  // addition,    concat,    concat,  concat
System.out.println(""   +     a     +      b);
                  // String+int  String+int
                  // concat,      concat

Allan · Answer 3 · 2019-03-18T07:50:17.320

First case:

System.out.println(a + b +" ");

In this order, a + b will first be evaluated as a int sum, then converted to a string when adding the space (the second + here is for string concatenation).

Second case:

System.out.println(a + b +" "+ a+b);

Here the first part will be int sum operation then converted to a string as we add the space (the 2nd + is for string concatenation), the rest will be string concatenation as the left operand is already a string.

Third case:

System.out.println(""+ a+b);

Same as 2nd.

Notes:

In order to change this behavior, just add parenthesis to force the int sum before the string concatenations.

System.out.println behavior when using String and int together

3 Answers3

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