When we create a 2d array such as int[][] a = new int[2][3] why is the resulting 2d array consist of a two-element array that contains three-element int arrays instead of the other way around. The reason why I'm confused is that when we make an array we do datatype[], so when we do int[2][3] why don't we put three int[2] arrays into an array with three spots (from the [3]).
The way it's implemented in Java is more logical. Consider the array element access expression: a[x][y]. Currently, it could be nicely decomposed to (a[x])[y] which means "we get an x-th element of a, then we get a y-th element of the result". So imagine if new int[2][3] produced an array of three elements, each is a two-element array. Then the x should be in range 0..2 and y should be in range 0..1 which is the opposite of the dimension order used at the array creation point. That would be absolutely confusing.
I guess you have a point with your logic. Eventhough you could also argument, writing int[2][3] means "first index can have 2 different values, second 3", what leads to the same as how it really works.
In the end, this is just a matter of specification and compilerbuilding. And since it is specified this way and not that way, it is implemented and works this way.
there
I have programmed in Java for about 4 years now. I was looking at a 2-D array and something caught my interest; the syntax.
So when declaring and initializing a 2-D array:
int [][] a = new int[2][3];
I know that I am creating an array of arrays of type int;
So how is type determined my first guess was that you have an array of type int [] i.e [] a is of type int []. I know that I am not expressing this well.
I'm trying to ask: Could someone explain the syntax in detail and explain which square bracket in declaration matches to which square bracket in initialization?
Sorry if my English is poor.
When you create a 2D array you are basically making an array in another array so by doing: int [][] a = new int[2][3]; You are declaring an array with a size of 2 containing arrays of integers with a size of 3. The number between the bracket indicates the size of both arrays.
When we write:
int iAmAnArray[][], iAmNotAnArray;
iAmAnArray = new int[2][3];
iAmNotAnArray = 13;
If you see the above code it clearly states that by using above style you can still declare normal variable along with the array variable, but when we write like bellow
int[][] iAmArray1,iAmArray2;
iAmArray1=new int[3][];
iAmArray2=new int[4][];
we can only declare array variable and one important thing is square bracket is not required for every variable, and is good approach when we have to declare or use too many variable of same type( say array).
I came across 2 examples from my notes which is about copying arrays.
The first example given below, stated that it is not the way to copy an array. But, when i tried to run the code, it managed to copy all the values from array1 to array2.
int []array1={2,4,6,8,10};
int []array2=array1;
for(int x:array2){
System.out.println(x);
}
The second example given, was saying the right way to copy an array.
int[] firstArray = {5, 10, 15, 20, 25 };
int[] secondArray = new int[5];
for (int i = 0; i < firstArray.length; i++)
secondArray[i] = firstArray[i];
My question is, are these 2 examples appropriate to be applied in coding or Example 2 is preferred. If you were my lecturer, I were to apply Example 1.. I will be given less mark compared to Example 2 method or just the same?
The first example doesn't copy anything. It assigns a reference of the original array to a new variable (array2), so both variables (array1 and array2) refer to the same array object.
The second example actually creates a second array and copies the contents of the original array to it.
There are other easier ways of copying arrays. You can use Arrays.copyOf or System.arraycopy instead of explicitly copying the elements of the array via a for loop.
int[] secondArray = Arrays.copyOf (firstArray, firstArray.length);
or
int[] secondArray = new int[firstArray.length];
System.arraycopy(firstArray, 0, secondArray, 0, firstArray.length);
Perfect way to copy elements is
System.arraycopy(array1,0, array2, 0, array1.length);
That above code is replacement for you second example which avoids a for loop.
And where in your first example, you are just referring the first array. So what ever changes happened to the first array can be seen from second array.
My question is, are these 2 examples appropriate to be applied in coding or Example 2 is preferred.
See again, they are not doing the something to compare. First one pointing to array reference and second snippet of code referencing elements it it. So you just can't compare them.
And there are other ways to copy array elements as well just like others mentioned and I prefer System.arraycopy because
1)Arrays.copyOf creates another array object internally and returns it where as System.arraycopy uses the passed array.
2) Clone is the too slow. Never use it unless you have no other choice.
There are few reasons I'm avoid clone. Here is
Josh Bloch analysis on clone vs copy constructor (when you clone individual elements).
To demonstrate the problem with the first method, try this:
int[] array1 = {2, 4, 6, 8, 10};
int[] array2 = array1;
array1[0] = 0;
System.out.println(array2[0]);
What do you think this will print?
0
It will print 0, because array2 now points to array1:
both variables now refer to the same object,
so modifying the content of any one of these will appear to modify both.
So your first method is NOT called copying an array.
It's a simple assignment, from one variable to another,
in this case assigning the value of array1 to array2,
so that both variables point to the same thing.
As for your second method,
here's a much simpler way to accomplish the same thing:
int[] array2 = array1.clone();
In your first example, you end up with only one array with two variables referring to it. Assignment only passes a reference to the array. So both firstArray and secondArray are pointing to the same array. It's not a copy. Try to set firstArray[0] = 99 and print secondArray and you'll see it's the same array.
In the second example, it's an actual copy - you have created a new array and copied each value. Now, assigning firstArray[0] = 99 won't change the array referred to by secondArray because it's a copy.
So the answer is: if you give the first way, you'll get lower marks.
In real use, there are better ways to copy arrays, using System.arraycopy, Arrays.copyOf or firstArray.clone().
int[] secondArray = new int[5];
In the second example, you are creating an array of int using new operator and copying the elements to it manually using for loop. Whenever you use new operator in java, a new memory allocation happens (object creation).
Where as in the first, it is just reference assignment, both are pointing to same array. You can easily verify that by changing the content of one array see the same effect into the second array. This will not happen in the second array copying.
int []array2=array1;
my question is really simple (which doesn't imply that the answer will be as simple.. :D )
why do arrays in C++ include the size as part of the type and Java's do not?
I know that Java array reference variables are just pointers to arrays on the heap,but so are C++ pointers to arrays,but I need to provide a size even then.
Let's analyze C++ first:
// in C++ :
// an array on the stack:
int array[*constexpr*];
// a bidimensional array on the stack:
int m_array[*constexpr1*][*constexpr2*];
// a multidimensional array on the stack:
int mm_array[*constexpr1*][*constexpr2*][*constexpr3*];
// a dynamic "array" on the heap:
int *array = new int[n];
// a dynamic bidimensional "array" on the heap:
int (*m_array)[*constexpr*] = new int[n][*constexpr*];
// a dynamic multidimensional "array" on the heap:
int (*mm_array)[*constexpr*][*constexpr*] = new int [n][*constexpr1*][*constexpr2*];
n doesn't have to be a compile time constant expression,all the elements are default initialized. Dynamically allocated "arrays" are not of type array,but the new expression yields a pointer to the first element.
So when I create a dynamic array,all dimensions apart the first one,must be constant expressions (otherwise I couldn't declare the pointer to hold their elements). Is it right??
Now to Java.I can only allocate array on the heap,since this is how Java works:
// a dynamic array on the heap:
int[] array = new int[n];
// a dynamic bidimensional array on the heap:
int[][] m_array = new int[n][];
// a dynamic multidimensional array on the heap:
int[][][] mm_array = new int [n][][];
In Java, it doesn't seem to care about array size when defining an array reference variable (it's an error in Java to explicitly provide a size),and so I just need to provide the size for the first dimension when creating the array. This allows me to create jagged array,which I'm not sure I can create in C++ (not arrays of pointers).
can someone explain me how's that? maybe what's happening behind the curtains should make it clear. Thanks.
That's because in Java, all arrays are single-dimensional. A two-dimensional array in Java is merely an array of references to one-dimensional arrays. A three-dimensional array in Java is merely a one-dimensional array of references to arrays of references to arrays of whatever base type you wanted.
Or in C++ speak, an array in Java, if it's not an array of primitive, it's an "array of pointers".
So, for example, this code:
int[][][] arr3D = new int [5][][];
System.out.println(Arrays.deepToString(arr3D));
Would yield the output:
[null, null, null, null, null]
You can decide to initialize one of its elements:
arr3D[2] = new int[3][];
And the output from the same println would now be:
[null, null, [null, null, null], null, null]
Still no ints here... Now we can add:
arr3D[2][2] = new int[7];
And now the result will be:
[null, null, [null, null, [0, 0, 0, 0, 0, 0, 0]], null, null]
So, you can see that this is an "array of pointers".
In C++, when you allocate a multi-dimensional array the way you described, you are allocating a contiguous array which actually holds all the dimensions of the array and is initialized all the way through to the ints. To be able to know whether it's a 10x10x10 array or a 100x10 array, you have to mention the sizes.
Further explanation
In C++, the declaration
int (*mm_array)[5][3];
means "mm_array is a pointer to a 5x3 array of integers". When you assign something to it, you expect that thing to be a pointer to a contiguous block of memory, which is at least big enough to contain 15 integers, or maybe an array of several such 5x3 arrays.
Suppose you didn't mention that "5" and "3".
int (*mm_array)[][]; // This is not a legal declaration in C++
Now, suppose you are handed a pointer to a newly allocated array, and we have statements like:
mm_array[1][1][1] = 2;
Or
mm_array++;
In order to know where to put the number, it needs to know where index 1 of the array is. Element 0 is easy - it's right at the pointer. But where is element 1? It's supposed to be 15 ints after that. But at compile time, you won't know that, because you didn't give the sizes. The same goes for the ++. If it doesn't know that each element of the array is 15 ints, how will it skip that many bytes?
Furthermore, when is it a 3x5 or a 5x3 array? If it needs to go to element mm_array[0][2][1], does it need to skip two rows of five elements, or two rows of three elements?
This is why it needs to know, at compile time, the size of its base array. Since the pointer has no information about sizes in it, and merely points to a contiguous block of integer, that information will need to be known in advance.
In Java, the situation is different. The array itself and its sub-arrays, are all Java objects. Each array is one-dimensional. When you have an expression like
arr3D[0][1][2]
arr3D is known to be a reference to an array. That array has length and type information, and one dimension of references. It can check whether 0 is a valid index, and dereference the 0th element, which is itself a reference to an array.
Which means that now it has type and length information again, and then a single dimension of references. It can check whether 1 is a valid index in that array. If it is, it can go to that element, and dereference it, and get the innermost array.
Since the arrays are not a contiguous block, but rather references to objects, you don't need to know sizes at compile time. Everything is allocated dynamically, and only the third level (in this case) has actual contiguous integers in it - only a single dimension, which does not require advance calculation.
I guess your real question is, why a stack array must have a fixed size at compile time.
Well, for one, that makes it easier to calculate the addresses of following local variables.
Dynamic size for stack array isn't impossible, it's just more complicated, as you would imagine.
C99 does support variable length arrays on stack. Some C++ compilers also support this feature. See also Array size at run time without dynamic allocation is allowed?
I believe this has to do with what code the compiler issues to address the array. For dynamic arrays you have an array of arrays and cells are addressed by redirecting a redirection.
But multidimensional arrays are stored in contiguous memory and the compiler indexes them using a mathematical formula to calculate the cell position based upon each of the array's dimensions.
Therefore the dimensions need to be known (declared) to the compiler (all except the last one).
Correction:
C sometimes has dimension
Java
Sometype some[];
declaration is itself an (declaration of) reference to Object and can be changed (to new instance or array). This may be one reason so in java dimension cannot be given "on the left side". Its near to
Sometype * some
in C (forgive me, array in Java is much more intelligent and safe)
if we think about pass array to C function, formal situation is similar like in Java. Not only we don't have dimension(s), but cannot have.
void func(Sometype arg[])
{
// in C totally unknown (without library / framework / convention etc)
// in Java formally not declared, can be get at runtime
}
In Java, it doesn't seem to care about array size when defining an array reference variable (it's an error in Java to explicitly provide a size),
It is not Java doesn't care about the initial array size when you define an array. The concept of an array in Java is almost totally different from C/C++.
First of all the syntax for creating an array in Java is already different.
The reason why you are still seeing C/C++ look-alike square brackets in Java when declaring arrays is because when Java was implemented, they tried to follow the syntax of C/C++ as much as possible.
From Java docs:
Like declarations for variables of other types, an array declaration has two components: the array's type and the array's name. An array's type is written as type[], where type is the data type of the contained elements; the brackets are special symbols indicating that this variable holds an array. The size of the array is not part of its type (which is why the brackets are empty)
When you declare an array in Java, for e.g.:
int[] array;
You are merely creating an object which Java called it an array (which acts like an array).
The brackets [ ] are merely symbol to indicate this is an Array object. How could you insert numbers into a specific symbol which Java uses it to create an Array Object!!
The brackets looks like what we used in C/C++ array declaration. But Java gives a different meaning to it to the syntax looks like C/C++.
Another description from Java docs:
Brackets are allowed in declarators as a nod to the tradition of C and C++.
Part of your question:
This allows me to create jagged array,which I'm not sure I can create in C++ (not arrays of pointers).
From Java Docs:
In the Java programming language, a multidimensional array is an array whose components are themselves arrays. This is unlike arrays in C or Fortran. A consequence of this is that the rows are allowed to vary in length
If you are interested to find out more on Java Arrays, visit:
here
here
and here
The difference between in arrays in C++ and Java is that Java arrays are references, like all non-primitive Java objects, while C++ arrays are not, like all C++ objects (yes, you hear a lot that C++ arrays are like pointers, but see below).
Declaring an array in C++ allocates memory for the array.
int a[2];
a[0] = 42;
a[1] = 64;
is perfectly legal. However, to allocate memory for the array you must know its size.
Declaring an array in Java does not allocate memory for the array, only for the reference, so if you do:
int[] a;
a[0] = 42;
you'll get a NullPointerException. You first have to construct the array (and also in Java, to construct the array you need to know its size):
int[] a = new int[2];
a[0] = 42;
a[1] = 64;
So what about C++ array being pointers? Well, they are pointers (because you can do pointer arithmetic with them) but they are constant pointers whose value is not actually stored in the program but known at compile time. For this reason the following C++ code will not compile:
int a[2];
int b[2];
a = b;
You're confusing the meaning of some of your C++ arrays:
e.g., your 'm_array' is a pointer to an array of values - see the following compilable C++ example:
int array_of_values[3] = { 1, 2, 3 };
int (*m_array)[3] = &array_of_values;
the equivalent Java is:
int[] array_of_values = {1, 2, 3};
int[] m_array = array_of_values;
similarly, your 'mm_array' is a pointer to an array of arrays:
int array_of_array_of_values[3][2] = { 1, 2, 3, 4, 5, 6 };
int (*mm_array)[3][2] = &array_of_array_of_values;
the equivalent Java is:
int[][] array_of_array_of_values = { {1, 2}, {3, 4}, {5, 6} };
int[][] mm_array = array_of_array_of_values;
How should I go about asking a 2-dimensional array how many rows it has?
Firstly, Java technically doesn't have 2-dimensional arrays: it has arrays of arrays. So in Java you can do this:
String arr[][] = new String[] {
new String[3],
new String[4],
new String[5]
};
The point I want to get across is the above is not rectangular (as a true 2D array would be).
So, your array of arrays, is it by columns then rows or rows then columns? If it is rows then columns then it's easy:
int rows = arr.length;
(from the above example).
If your array is columns then rows then you've got a problem. You can do this:
int rows = arr[0].length;
but this could fail for a number of reasons:
The array must be size 0 in which case you will get an exception; and
You are assuming the length of the first array element is the number of rows. This is not necessarily correct as the example above shows.
Arrays are a crude tool. If you want a true 2D object I strongly suggest you find or write a class that behaves in the correct way.
Object[][] data = ...
System.out.println(data.length); // number of rows
System.out.println(data[0].length); // number of columns in first row
int[][] ia = new int[5][6];
System.out.println(ia.length);
System.out.println(ia[0].length);
It depends what you mean by "how many rows".
For a start, a 2-dimensional array is actually a 1-D array of 1-D arrays in Java. And there is no requirement that a 2-D array is actually rectangular, or even that all elements in the first dimension are populated.
If you want to find the number of elements in the first dimension, the answer is simply array.length.
If you want to find the number of elements in the second dimension of a rectangular 2-D array, the answer is `array[0].length.
If you want to find the number of elements in the second dimension of a non-rectangular or sparse 2-D array, the answer is undefined.