Java parseInt() question

Question

Why is it that when I use parseInt for this:

private String certainNumber;

public int getNumber()
{
    return Integer.parseInt(certainNumber);
}

It compiles.

But If I were to do this:

public String getStreetNumber()
{
    return streetNumber;
}

and parseInt the returned value like so:

@Override
public int compareTo(Object o) 
{
    Address tempAddress = (Address)o;
    if(Integer.parseInt(getStreetNumber()) < tempAddress.Integer.parseInt(getStreetNumber()))
    {
        return -1;
    }
... // etc.
}

It does not compile?

edit: tried the suggestions... still not compiling?

edit2: Thanks for the help guys!

Beacause you are calling parseInt directly, You must use Integer.parseInt() as it it is an static method of Integer class — TeaCupApp
– TeaCupApp, Commented Aug 9, 2011 at 5:36
If you wanted to, you could statically import the parseInt method. — Hugh
– Hugh, Commented Aug 9, 2011 at 5:37

cwallenpoole · Accepted Answer · 2011-08-09 05:36:43Z

4

That is because parseInt is a method which belongs to the Integer class: you have to call it Integer.parseInt(value);. I highly doubt that you have a parseInt function in either your custom class (I suspect that this is all part of an Address class?) or the tempAddress instance.

Try this:

public int compareTo(Object o) 
{
    Address tempAddress = (Address)o;
    if(Integer.parseInt(getStreetNumber()) < 
          // you need to parse the return value of tempAddress's getStreetNumber() 
          // not get the tempAddress's parseInt of this.getStreetNumber()
          Integer.parseInt(tempAddress.getStreetNumber()))
    {
        return -1;
    }
 // etc...
}

answered Aug 9, 2011 at 5:36

cwallenpoole

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Comments

Eng.Fouad · Accepted Answer · 2011-08-09 05:35:18Z

3

Because you called parseInt() not Integer.parseInt()

answered Aug 9, 2011 at 5:35

Eng.Fouad

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Comments

Nivas · Accepted Answer · 2011-08-09 05:40:40Z

2

parseInt is a static method in the Integer class. You call it as Integer.parseInt(the_string_you_want_to_parse)

If you want to parse an string returned by a method in another class you call it as Integer.parseInt(tempAddress.getStreetNumber())

answered Aug 9, 2011 at 5:40

Nivas

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Comments

Mark Peters · Accepted Answer · 2011-08-09 05:40:31Z

1

Others answered your direct question, but let me say that this is a very clean way of doing the same thing you're trying to do:

return Integer.valueOf(getStreetNumber())
    .compareTo(
       Integer.valueOf(tempAddress.getStreetNumber()));

Integer already implements Comparable, so you might as well leverage it.

answered Aug 9, 2011 at 5:40

Mark Peters

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Comments

RiaD · Accepted Answer · 2011-08-09 05:36:05Z

0

if(Integer.parseInt(getStreetNumber())

answered Aug 9, 2011 at 5:36

RiaD

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Comments

Sathwick · Accepted Answer · 2011-08-09 05:36:57Z

0

How can you call a parseInt on a Address reference???

> if(parseInt(getStreetNumber()) < tempAddress.parseInt(getStreetNumber()))

Also you should always say Integer.parseInt(), you cant call the parseInt directly.

answered Aug 9, 2011 at 5:36

Sathwick

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Comments

Maurice Perry · Accepted Answer · 2011-08-09 05:38:49Z

0

Yes, you can also statically import it:

import static java.lang.Integer.parseInt;

answered Aug 9, 2011 at 5:38

Maurice Perry

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