I know array lookup has O(1) time, so it cannot be looping through. Does the program store the memory locations of the indices of the array, or how does it look the index instantaneously?
Array elements are always spaced at equal distances in the memory, so finding an element given an index requires a multiplication by the size of the element and an addition of the array's base in memory. Both operations are often done within the space of a single instruction in hardware by employing an appropriate addressing mode.
underneath...
its a memory address + (index postion * the size of the things in the array)
Try this,
1. Arrays are consecutive memory locations which are stored in Heap, as Arrays are objects in java.
2. Assume i have an Array of String as an instance variable
String[] arr = {1,2,3,4,5};
Now its like this
arr[0] = 1
arr[1] = 2
arr[2] = 3
arr[3] = 4
arr[4] = 5
{1,2,3,4,5} are stored over the heap, and Considering array "arr" as instance variable, will lives within the object on the heap.
Now arr will hold the address of the very first element of the Array that is 1. "arr" which is an object reference array variable, will be inside the object, and {1,2,3,4,5} outside somewhere on the heap.
Array elements are stored in a consecutive block, if they grow they need to be moved to a new place. The elements are then accessed using an offset from where the array begins.
In C you can access the element of index i in an array called a using two different methods:
int arrayElement = a[i];
int arrayElement = (int)(a + i * sizeof(int));
This is more or less how it is done in Java under the hood.
Related
Hello I am research about that, but I cannot found anything in the oracle website.
The question is the next.
If you are using an static Array like this
int[] foo = new int[10];
And you want add some value to the 4 position of this ways
foor[4] = 4;
That don't shift the elements of the array so the time complexity will be O(1) because if you array start at 0x000001, and have 10 spaces, and you want put some in the x position you can access by (x*sizeOf(int))+initialMemoryPosition (this is a pseudocode)
Is this right, is this the way of that this type of array works in java, and if its time complexity O(1)
Thanks
The question is based on a misconception: in Java, you can't add elements to an array.
An array gets allocated once, initially, with a predefined number of entries. It is not possible to change that number later on.
In other words:
int a[] = new int[5];
a[4] = 5;
doesn't add anything. It just sets a value in memory.
So, if at all, we could say that we have somehow "O(1)" for accessing an address in memory, as nothing related to arrays depends on the number of entries.
Note: if you ask about ArrayList, things are different, as here adding to the end of the array can cause the creation of a new, larger (underlying) array, and moving of data.
An array is somewhere in memory. You don’t have control where, and you should not care where it is. The array is initialized when using the new type[size] syntax is used.
Accessing the array is done using the [] index operator. It will never modify size or order. Just the indexed location if you assign to it.
See also https://www.w3schools.com/java/java_arrays.asp
The time complexity is already correctly commented on. But that is the concern after getting the syntax right.
An old post regarding time complexity of collections can be found here.
Yes, it takes O(1) time. When you initialize an array, lets say, int[] foo = new int[10],
then it will create a new array with 0s. Since int has 4 bytes, which is 32 bits, every time assign a value to one element, i.e., foo[4] = 5, it will do foo[32 x input(which is 4)] = value(5); That's why array is 0-indexed, and how they assign values in O(1) time.
Can I Use Array Variables With Changed Length Each Time
like
int [] primes = new int[6];
...
primes = new int[4];
Can I Use Prime Again Like I Did Above ? Is It Possible ? If Not Can You Give Me Reason Thanks .
You are not reusing the same array, but creating new instances of array object. The java array object has a fixed length (final int), so it's impossible to change the length of an array after instantiation.
Some explanation to your question.
You can run the following code and will get better understanding.
As java can't get the address, so here using hashcode to show the object address.
int[] prime = new int[]{1,2,3,4};
System.out.println(Arrays.hashCode(prime));
System.out.println("prime[0]="+prime[0]);
prime = new int[]{4,3,2,1,0};
System.out.println(Arrays.hashCode(prime));
System.out.println("prime[0]="+prime[0]);
Output like this:
955331
prime[0]=1
1045631
prime[0]=4
Array can not be resized once memory is being allocated as array get a contiguous memory block.
int arr[] = new int[3];
Once memory space is hold, resizing the array is not possible as next contiguous memory block may not be available. You can either create a new array and shift all the elements of previous array to the new array with different size.
I would recommend using ArrayList for dynamic array. If you get in any such situation.
Two separate arrays
You are not changing the length of the array. You are creating a second array, separate and distinct from the first.
You created two arrays but used only a single reference variable (primes) to track them. When you assigned the second array to primes, the first was forgotten. The first array is still floating around in memory, a candidate for eventual garbage collection if no other reference points to it.
ArrayList
If you want resizing, use a List implementation such as ArrayList rather than a mere array. You’ll need to store objects (Integer) rather than primitives (int). The auto-boxing feature in Java can mask the difference.
int initialCapacity = 6 ;
ArrayList< Integer > primes = new ArrayList<>( initialCapacity ) ;
primes.add( 7 ) ; // The primitive `int` 7 literal is automatically converted via autoboxing to be an `Integer` Object.
The ArrayList automatically grows its capacity as needed to store additional items.
If you wish to free up extra capacity not allocated to hold any objects, call ArrayList::trimToSize().
int a[][]=new int[2][]; // It works without any error
Why is the second dimension missing in this snippet?
A 2D array is, technically, an array of arrays. The code that you have specified tells you how many arrays you want to have.
You can further initialize this as follows:
int a[][] = new int[2][];
a[0] = new int[3];
a[1] = new int[5];
Something like new int[2][2] is nothing more than a condensed version of the code above.
It's not mandatory because the second dimension is not required to calculate how much memory is required to hold the array.
Compare the following:
int[] a = new int[2];
In this case the JVM needs to be instructed to allocate space for one array that holds two integers.
On the other hand:
int[][] = new int[2][];
In this case the JVM needs to be instructed to allocate space for two references to integer array objects. It doesn't matter what size these integer array objects end up being, as it doesn't change the size of the reference.
In fact, these two arrays can have different sizes or even not be created at all.
Second dimension in array is optional in Java. You can create a two dimensional array without specifying both dimension e.g. int[4][] is valid array declaration.
The reason behind that is Java doesn't support multi-dimensional array in true sense. In a true two dimensional array all the elements of array occupy a contiguous block of memory , but that's not true in Java.
Instead a multi-dimensional array is an array of array. For example two dimensional array in Java is simply an array of one dimensional array like String[][] is an array of array of String[] or "array of array of strings". This diagram shows how exactly two dimensional arrays are stored in Java :
I was just playing around with JavaScript and got stuck with a simple program.
I declared an array in JavaScript like
var a = [0, 1, 2];
Then as there is no fixed size for an array in JavaScript and we can add more to the array, I added another integer to array.
a[3] = 3;
And as expected If I try to access a[4] I am definitely going to get it as undefined.
Now, if I take an array
var a = [0,1,2];
And add another element
a[4] = 4;
I have intentionally not defined a[3], and this also gives me a[3] as undefined.
Here is a fiddle where this can be observed: http://jsfiddle.net/ZUrvM/
Now, if I try the same thing in Java,
int[] a = new int[4];
a[0] = 0;
a[1] = 1;
a[3] = 3;
Then I end up with
a[2] = 0;
You can see this on ideone: https://ideone.com/WKn6Rf
The reason for this in Java I found is that the four variables are defined while declaring the array and we can only assign values to the declared size of array.
But in JavaScript when I declare an array of size 3 and then add 5th element why does it not consider the 4th element to be null or 0 if we have increased the array size beyond 4?
Why do I see this strange behavior in JavaScript, but not in other languages?
Why is this strange behavior in JavaScript?
Because arrays are only objects. If you access a nonexisting property, you get back undefined. You simply didn't assign an element at index 3, so there is nothing.
Auto-growing the array by assigning higher indices does not change this behaviour. It will affect the .length property, yes, but the intermediate indices will stay nonexistent. This is known as a sparse array.
Why is this strange behaviour in Java / C / C++?
Because arrays are chunks of allocated memory, and when allocating an array of size 4, all its elements take their values from that memory location. To avoid indeterminate values, in some languages/occasions the fields get default-initialised, typically with 0.
JavaScript isn't a strongly typed language.
In Java you declared an array of integers. The default value of any element in that array is 0.
In JavaScript, when you declare an array, the default value is undefined as it can hold anything, number, string, any object (including another array). All elements of a JavaScript array don't have to be the same type.
An array is a continuous collection of data. Say if you have a value at index 1 and index 10, the rest will be filled by undefined.
You can't create an array with empty values. 'Empty in your sense', is not undefined or null, it's empty :)
That continuous data is undefined means default assignment in JavaScript.
Say if you defined a variable,
var X;
console.log(X) //Undefined
it's not null. null is a user explicitly specifying a way to tell there is an empty data, and undefined is the JavaScript engine way.
I'm a Java noob. I don't know very much about the language (at least, not enough to do complex things) right now, but I'm getting there!
I know you can test the length of a single-dimensioned array by doing arr.length, but is it possible to test other dimensions (in a multidimensional array)?
Yes, it is. Obviously, to test the first dimension you would just do arr.length. Subsequent dimensions are tested by using length with the [0] element of that particular dimension. For example, consider this array:
int[][][][] arr = new int[10][11][12][13];
To test the...
first dimension: arr.length;
second dimension: arr[0].length;
third dimension: arr.[0][0].length;
fourth dimension: arr.[0][0][0].length;
A multidimensional array is just an array of arrays, and each array in the array can have different lengths. I.e.:
int arr[][] = new int[2][];
arr[0] = new int[5];
arr[1] = new int[10];
System.out.println(arr.length);
System.out.println(arr[0].length);
System.out.println(arr[1].length);
Now you have a two-dimensional array. The first dimension (outer) can be referred to as arr and is of size 2. The inner arrays can be referred to as arr[0] and arr[1] and have lengths 5 and 10, respectively. Since each of these refers to a normal Java array, you can use all the normal ways of accessing an arry on them. Further, since we create multidimensional arrays by putting arrays in arrays, you can have as many dimensions as you want, and you access each further level down by indexing into the one above: arr[2][1][5][11][3][0][123][5][42][9][7]....length
Given, say, a 2D array, you can access the length of the ith inner array with arr[i].length.
If you haven't already seen it, check out Arrays (The Java™ Tutorials).
You can test the other dimensions of the other dimensions by directly referrencing the dimension you are wanting to test.
For example, if you have a 2 dimensional array with 3 items in the primary dimension then you can identify the length of each of them by using arr[0].length, arr[1].length, and arr[2].length.
the code below just sets up an array, and then verifies that the lengths are what we expect them to be.
public void testLength(){
//setup the array you are wanting to test
int[][] foo = new int[2][5];
// this just makes sure that we do get 2 dimensions within the primary
Assert.assertEquals(2,foo.length);
Assert.assertEquals(5,foo[0].length);
// change the array stored within foo[0]
foo[0]= new int[8];
Assert.assertEquals(8,foo[0].length);
}