1. Java pass by value or pass by reference
Java is ALWAYS pass-by-value (meaning make another new copy ).
That’s why when we change object’s value within function, the original object’s value is untouchable.
public class Main
{
public static void main (String[] args)
{
Foo f = new Foo(“f”);
changeReference(f);
modifyReference(f);
}
}
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public class Main
{
public static void main (String[] args)
{
Foo f = new Foo(“f”);
changeReference(f);
modifyReference(f);
}
public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
}
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public class Main
{
public static void main (String[] args)
{
Foo f = new Foo(“f”);
changeReference(f);
modifyReference(f);
}
public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
}
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public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
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public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
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public class Main
{
public static void main (String[] args)
{
Foo f = new Foo(“f”);
changeReference(f);
modifyReference(f);
}
public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
public static void modifyReference (Foo c)
{
c.setAttribute(“c”);
}
}
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public class Main
{
public static void main (String[] args)
{
Foo f = new Foo(“f”);
changeReference(f);
modifyReference(f);
}
public static void changeReference (Foo a)
{
Foo b = new Foo(“b”);
a=b;
}
public static void modifyReference (Foo c)
{
c.setAttribute(“c”);
}
}
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Reference:
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