Instantiating a value type

To explain your question, look at the example below.

struct Person
{
    public Person(int age)
    {
        this.Age = age + 10;
    }
    public int Age;
}

And then access it like

Person person =new Person();
person.Age = 10;

Person person1 = new Person(10);

What you expect Person's age is 10, but Person1's value will be 20. Using the new key word you can call a specific constructor. For your second case you are calling a default construtor which assign a 0 to the variable and then you are assigning value to it. In the first case the int will accept a compatible value which implicitly to the value type.

For the int, the both ways you are calling are same, because you are using simply a default constructor.

Hope this helps.

Note: value types are different from reference types In the above example if the person is a class then

Person1= Person; Person1.Age=30; then Person.Age also equals 30. But in case of value type Person.Age will have the original value. Value types stores on the stack and a new variable will be in a new memory address. Reference variable pointers are in the stack which points to an address in the heap. So two variable can hold a common address in the heap.

Good luck

see this link

http://www.coderanch.com/t/468841/java/java/difference-between-int-maxrows-Integer[^]

Apart from what Albein already said,
When you use a new expression with a structure, the object is intialised to Zero (and nulls if the structure include fields that were reference types).

So when you write this -

int a = new int();

a is implicitly initialised to 0.
Consider this example,

int a;
a = a + 10;

This will give you a compiler error - use of unassigned local variable 'a'

But for the code below,

int a = new int();
a = a + 10;

It works fine. As a was by default assigned 0.

Hope this helps as well!