Java chained constructor calling error [duplicate]

Question

This made me mad (Eclipse Kepler)

public class FastReader 
{
public static void main (String[] args)
{
    FastReader a = new FastReader("hi");
}
public FastReader(int a)
{

}
public FastReader(String b)
{
    FastReader(10);
}
}

And I get this error:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
The method FastReader(int) is undefined for the type FastReader    
at FastReader.<init>(FastReader.java:14)
at FastReader.main(FastReader.java:6)

It almost made me mad! help me get rid of this! Thank you!

Google is your friend. Googling for "chain constructors in java" shows the answer as its very first link. — JB Nizet
– JB Nizet, Commented Aug 21, 2013 at 13:43
@JBNizet nowhere in the exception is someone able to deduce that they need to look for "chaining constructors". I think this question is very valid, a duplicate for sure, but valid — Isaac
– Isaac, Commented Sep 5, 2017 at 23:41
@Isaac I never said the question was invalid. I just said thet, since the OP wants to call a constructor from another constructor in Java, the first thing to do is to do a tiny bit of research by googling, for example "call a constructor from another constructor in Java". Just doing that would have led to the answer immediately. — JB Nizet
– JB Nizet, Commented Sep 6, 2017 at 5:22

Reimeus · Accepted Answer · 2013-08-21 13:41:12Z

12

Use

public FastReader(String b) {
   this(10);
}

answered Aug 21, 2013 at 13:41

Reimeus

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2 Comments

Alireza Mohamadi Over a year ago

You say that my way of calling constructors is completely wrong bro? thanks

Reimeus Over a year ago

Yes, the compiler is telling you this, the above is the correct syntax for constructor chaining

Santosh Jagannath · Accepted Answer · 2013-08-21 14:37:47Z

2

public FastReader(String b)
{
    this(10);
}

This is the correct way to call the same class constructor. If you want to call a same class constructor use the keyword 'this' if you want call the parent class constructor use the keyword 'super'.

answered Aug 21, 2013 at 14:37

Santosh Jagannath

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