Why this output is generated by the following code, why not error/exception is generated?

Question

class VarAgs{

    // variable length parameters
    static void vaTest(String ... str){
        System.out.println(str.length + " contains : ");
        for(String s:str){  
            System.out.print(s+" ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        String s1[] = {"hi","hello"};
        vaTest(s1);
        vaTest(s1 + " 1 ");
    }
}

Output:

2 contains : 
hi hello 
1 contains : 
[Ljava.lang.String;@2a139a55 1

Welcome to Stack Overflow. What exception would you expect, and why? Note that in the second call to `vaTest` you're passing a single string, so it's equivalent to `vaTest(new String[] { s1 + " 1 " })` — Jon Skeet, Mar 05 '19 at 07:57

score 0 · Answer 1 · answered Mar 05 '19 at 08:00

When you concatenate an array with a string in

    vaTest(s1 + " 1 ");

Then the reference to the array is converted into a string

 [Ljava.lang.String;@2a139a55

So the method vaTest will receive the

[Ljava.lang.String;@2a139a55 1

as an array with size of 1.

amseager · Answer 2 · 2019-03-05T20:49:36.893

0

When you do vaTest(s1 + " 1 ");, a string representation of s1 would be used because of string concatenation rules ("+" sign). So instead of a formatted array you get some kind of that: [Ljava.lang.String;@2a139a55

Replace it with vaTest(Arrays.toString(s1) + " 1 "); to get expected result.

edited Mar 05 '19 at 20:49

answered Mar 05 '19 at 08:00

amseager

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score 0 · Answer 3 · answered Mar 05 '19 at 08:00

It works as following:

s1 is implicitly converted to String (s1.toString()), this results in [Ljava.lang.String;@2a139a55.
Next, string 1 is added to the above string. This results in [Ljava.lang.String;@2a139a55 1
Next, this resulting string is passed inside single-element array of type String[] into varargs() function.

Why this output is generated by the following code, why not error/exception is generated?

3 Answers3