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;
Related
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.
I was under the impression that storing and accessing data beyond the upper bound specified for arrays will result in Exception but I was coding a solution and I stumbled upon a scenario where the array stored more elements than the upper bound.
public static void main(String[] args) {
int[][] arr = new int[1][5];
String values = "1 2 3 4 5 6";
arr[0] = Arrays.asList(values.split(" ")).stream().mapToInt(Integer::parseInt).toArray();
for (int i : arr[0]) {
System.out.println(i);
}
}
The above program outputs 1 2 3 4 5 6 in separate lines while clearly I have specified my upperbound as 5. Can someone explain this weird phenomenon?
Can someone explain this weird phenomenon?
Here goes:
int[][] arr = new int[1][5];
is declaring an int[][] and initializing it to a structure consisting of one top-level int[][] of length 1, and one second level int[] subarray of length 5. The first cell of the former refers to the latter.
Then:
arr[0] = Arrays.asList(values.split(" "))
.stream()
.mapToInt(Integer::parseInt)
.toArray();
is creating an int[] of length 6 and assigning it to arr[0]. This replaces the int[] subarray of length 5 that was created in the earlier initialization.
The intuitive behavior is that the "shape" of arr changes from a int[1][5] to int[1][6] when you do that assignment.
You then commented:
"Ohh basically then it treats as an array initialization rather than an assignment then."
Strictly, no. If you think of arr[0] = ... as an array initialization, it gets confusing. It is actually an assignment of a newly create array of type int[] to a cell of an existing "array of int[]".
This may sound like a fine distinction, but it goes to the heart of the matter. If you read the JLS and the JVM specs deeply, you will see that Java array types and objects are fundamentally one dimensional. Thus int[][] is a syntactic shorthand1 for "array of int[]", and arr[i][j] is a shorthand for (arr[i])[j].
Thus when we do arr[0] = ... in the example, we are not initializing something. Instead, we are changing a data structure that was initialized previously. At least, that is the Java language perspective.
From the perspective of the class where this is happening, this could be getting something into the initial state that is required by the class semantics. Thus it is reasonable to call that "initialization" ... from that perspective.
1 - This shorthand is so ingrained in the language that there isn't any other way to write the type "array of int[]" using Java syntax. The syntax for array types comes from C via C++, though the particular Java semantics of arrays doesn't.
After assignment, arr[0] is holding a list object that has bounds set to whatever Arrays.asList defines.
As I know, C++ stores array by putting 2D array values on a block of memory (continuous virtual memory?), which are fast for accessing value by index.
I came out this question after reading this, "using nested array to store 2D grid is efficient in C/C++, but in Java or other memory-managed languages, doing that will actually give you an array of rows where each element is a reference to the array of columns, which may not be as memory-friendly as you'd like".
Does "a reference to the array of columns" mean they actually be stored in many tiny blocks on memory?
Update
Sorry my question should be "If Java store 2D array on many tiny blocks, how is this easy for 'memory-management'"?
In an MxN matrix, it has M references for N arrays. And that is the reason that in C you must tell the second dimension when you want to pass an array as a function argument, and in Java you dont have to.
int[] []x[];
I know that
int[] x;
int x[];
int []x;
and similar declares an array perfectly fine, but how does 3 sets for brackets, work exactly?
It is legal because Java Language Specification - 10.2. Array Variables allows it:
The array type of a variable depends on the bracket pairs that may appear as part of the type at the beginning of a variable declaration, or as part of the declarator for the variable, or both.
(emphasis mine)
So to put it simply int[] []x[]; is same as int[][][] x;.
Generally we should avoid mixed way of placing [] (or even placing [] after variable name). IMO it is because in Java we are used to declaring variables as
FullTypeDescription variableName
instead of
SomeTypeInfo variable RestOfTypeInformation
and number of arrays dimensions is part of type information, so it should be separated.
Mixed type is allowed to let us write code like (to let C or C++ programmers migrate to Java easier):
int[][] a, b[];
instead of
int[][] a;
int[][][] b;
but still last way is preferred.
Although the follow 3 mean the same, the first one is the recommended one, since the type of x is an array of integers.
int[] x; // Recommended
int x[];
int []x;
So, the other one should be:
int[][][] x;
Which is an array of array of array of integers.
Depending on how it is used, it can be thought of as a 3-dimensional array, e.g. see Fundamentals of Multidimensional Arrays
Of course, in reality, Java doesn't have multi-dimensional arrays. It is, as first stated, an array of array of array of integers, and each sub-array may have different lengths. But that is a longer topic.
Here's an example:
int[][] array = new int[1][2];
Above we have a double-dimensional array. Really, it's an array of arrays. So declaring it like so:
int[] array[] = new int[1][2];
You are still declaring an array of arrays, aka a double-dimensional array. this means that this:
int[] []array[] = new int[1][2][3];
Is the same as this:
int[][][] array = new int[1][2][3];
Because brackets after the array identifier ('array') were adapted for C programmers, they are treated as though they're located before the identifier.
Hope this helps! :)
This question already has answers here:
Why does the indexing start with zero in 'C'?
(16 answers)
Closed 9 years ago.
Is there any special reason? I know that that's how the language has been written, but can't we change it?
And what are the challenges we'd face if the index start with 1?
For historical reasons, and reasons connected to how an array is "made" in memory.
In C, an array is a piece of memory (with some information at compiler level on its size). You have a pointer (a reference) to its first element. To go to its second element you can do
int array[10]; // your array
int *p = array; // reference to first element of array
int *q = p + 1; // reference to second element of array
int *r = p + 2; // reference to third element of array
clearly, for symmetry:
array[0] // reference to first element of array
array[1] // reference to second element of array
array[2] // reference to third element of array
the [x] operator of C is in fact compiled as array + x.
You see? Array are "base 0" in C. And for this reason in many other languages it's the same.
Now, C++ has its roots in C, Java has its roots in C++ and other languages, C# has its roots in C++, java and other languages... Same tree.
Basic clearly is another tree :-)
The basic reason behind it is that the computer remembers the address at wich the first part of any variable/object is stored. So the index represents the "distance" in between that and what you're looking for, so the first one is 0 away, the second 1...
In C and C++, array indexing is syntactic sugar for dereferencing an offset pointer. That is,
array[i] is equivalent to *(array + i). It makes sense for pointers to point to the beginning of their block of memory, and this implies that the first element of the array needs to be *array, which is just array[0].
Good question. Actually, almost all programming language implementations start indexing at 0.
While you could potentially use workarounds to make it seem like they're starting with 1, don't.
Dijkstra's article will give a much better defense of zero based arrays than I can:
http://www.cs.utexas.edu/users/EWD/transcriptions/EWD08xx/EWD831.html
One thing is that, you can reference and navigate the arrays using pointers. Infact the the array operations decay to pointer arithmetic at the back end.
Suppose you want to reach a nth element of an array then you can simply do (a + n) where a is the base address of an array(1-dimension), but if the subscript starts at 1 then to reach the nth element you would have to do (a + n -1) all the time.
This is because just by taking the name of an array you the the address of the starting element of it, which is the simplest way!