A static method parseInt(char[]) that converts an array of numeric characters to an int value

Question

Here's my code

public static int[] parseInt(char[] myChar)
{      
    int[] myInt = new int[myChar.length];
    for(int x = 0; x < myChar.length; x++)
    {
                myInt[x] = (int)myChar[x];
    }
    return myInt;
}

When I System.out.println(object.parseInt(myChar)); with values {'1','2','3'} I would expect it to print myInt with the corresponding unicode values for {'1','2','3'}.

Instead I get [I@15db9742

Any ideas on what I'm doing wrong?

Because you are getting int[] as results which is object. Loop over the results and print individual element. — kosa, Oct 02 '15 at 20:42
Your method does not convert to an int value. It converts to an array of integers. Your title seems to indicate that for an array `[1,2,3]` your are supposed to return the number 123. — RealSkeptic, Oct 02 '15 at 20:44

score 1 · Answer 1 · edited Oct 02 '15 at 20:46

1

That because you are printing the object, hence java shows:

[I - the name of the type (in this case it is one dimensional [ array of int) 
@ - joins the string together
15db9742the hashcode of the object.

the memory address by default. You can use:

System.out.println(Arrays.asList(parseInt(array)));

edited Oct 02 '15 at 20:46

Pshemo

122,468
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answered Oct 02 '15 at 20:42

Federico Piazza

30,085
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1

It's not the memory address. It's a concatenation of the object class name and its hash code. – RealSkeptic Oct 02 '15 at 20:43
1

"are see the memory address" that is not true. We are seeing hexadecimal hashcode. This value can be based on memory location, but doesn't reflect it (object can be moved in memory...) – Pshemo Oct 02 '15 at 20:43
@RealSkeptic very true, I'll update it – Federico Piazza Oct 02 '15 at 20:44

A static method parseInt(char[]) that converts an array of numeric characters to an int value

1 Answers1