How can I determine size of an array (length / number of items) in C#?
If it's a one-dimensional array a,
a.Length
will give the number of elements of a.
If b is a rectangular multi-dimensional array (for example, int[,] b = new int[3, 5];)
b.Rank
will give the number of dimensions (2) and
b.GetLength(dimensionIndex)
will get the length of any given dimension (0-based indexing for the dimensions - so b.GetLength(0) is 3 and b.GetLength(1) is 5).
See System.Array documentation for more info.
As #Lucero points out in the comments, there is a concept of a "jagged array", which is really nothing more than a single-dimensional array of (typically single-dimensional) arrays.
For example, one could have the following:
int[][] c = new int[3][];
c[0] = new int[] {1, 2, 3};
c[1] = new int[] {3, 14};
c[2] = new int[] {1, 1, 2, 3, 5, 8, 13};
Note that the 3 members of c all have different lengths.
In this case, as before c.Length will indicate the number of elements of c, (3) and c[0].Length, c[1].Length, and c[2].Length will be 3, 2, and 7, respectively.
You can look at the documentation for Array to find out the answer to this question.
In this particular case you probably need Length:
int sizeOfArray = array.Length;
But since this is such a basic question and you no doubt have many more like this, rather than just telling you the answer I'd rather tell you how to find the answer yourself.
Visual Studio Intellisense
When you type the name of a variable and press the . key it shows you a list of all the methods, properties, events, etc. available on that object. When you highlight a member it gives you a brief description of what it does.
Press F1
If you find a method or property that might do what you want but you're not sure, you can move the cursor over it and press F1 to get help. Here you get a much more detailed description plus links to related information.
Search
The search terms size of array in C# gives many links that tells you the answer to your question and much more. One of the most important skills a programmer must learn is how to find information. It is often faster to find the answer yourself, especially if the same question has been asked before.
Use a tutorial
If you are just beginning to learn C# you will find it easier to follow a tutorial. I can recommend the C# tutorials on MSDN. If you want a book, I'd recommend Essential C#.
Stack Overflow
If you're not able to find the answer on your own, please feel free to post the question on Stack Overflow. But we appreciate it if you show that you have taken the effort to find the answer yourself first.
for 1 dimensional array
int[] listItems = new int[] {2,4,8};
int length = listItems.Length;
for multidimensional array
int length = listItems.Rank;
To get the size of 1 dimension
int length = listItems.GetLength(0);
yourArray.Length :)
With the Length property.
int[] foo = new int[10];
int n = foo.Length; // n == 10
For a single dimension array, you use the Length property:
int size = theArray.Length;
For multiple dimension arrays the Length property returns the total number of items in the array. You can use the GetLength method to get the size of one of the dimensions:
int size0 = theArray.GetLength(0);
In most of the general cases 'Length' and 'Count' are used.
Array:
int[] myArray = new int[size];
int noOfElements = myArray.Length;
Typed List Array:
List <int> myArray = new List<int>();
int noOfElements = myArray.Count;
it goes like this:
1D:
type[] name=new type[size] //or =new type[]{.....elements...}
2D:
type[][]name=new type[size][] //second brackets are emtpy
then as you use this array :
name[i]=new type[size_of_sec.Dim]
or You can declare something like a matrix
type[ , ] name=new type [size1,size2]
What has been missed so far is what I suddenly was irritated about:
How do I know the amount of items inside the array? Is .Length equal .Count of a List?
The answer is: the amount of items of type X which have been put into an array of type X created with new X[number] you have to carry yourself!
Eg. using a counter: int countItemsInArray = 0 and countItemsInArray++ for every assignment to your array.
(The array just created with new X[number] has all space for number items (references) of type X already allocated, you can assign to any place inside as your first assignment, for example (if number = 100 and the variable name = a) a[50] = new X();.
I don't know whether C# specifies the initial value of each place inside an array upon creation, if it doesn't or the initial value you cannot compare to (because it might be a value you yourself have put into the array), you would have to track which places inside the array you already assigned to too if you don't assign sequentially starting from 0 (in which case all places smaller than countItemsInArray would be assigned to).)
In your question size of an array (length / number of items) depending on whether / is meant to stand for "alternative" or "divide by" the latter still has to be covered (the "number of items" I just gave as "amount of items" and others gave .Length which corresponds to the value of number in my code above):
C# has a sizeof operator (https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/operators/sizeof). It's safe to use for built-in types (such as int) (and only operates on types (not variables)). Thus the size of an array b of type int in bytes would be b.Length * sizeof(int).
(Due to all space of an array already being allocated on creation, like mentioned above, and sizeof only working on types, no code like sizeof(variable)/sizeof(type) would work or yield the amount of items without tracking.)
Related
I'm trying to make a sorting algorithm and asking the user to input the array size is a must. I am a beginner in C# so I don't have any idea how to do that.
This is the idea that came to my mind, but I'm having an error.
Console.WriteLine("Enter how many elements you want to be sorted:");
a = Convert.ToInt32(Console.ReadLine());
int[] MyArray= new int[a] {""};
Visual Studio says that 'a constant value is expected'. How could I make the array length a ReadLine? My goal is for the user to decide which array length they want the program to show and that the elements inside the array would be system generated based on the array length that the user chose.
You can initialise an array like this:
int[] MyArray= new int[a];
But, I would also point out, that you could use a dynamic collection (such as a list), then you don't need to ask up front how many items, you just keep adding items until the user decides to stop.
ICollection<int> myCollection = new List<int>();
myCollection.Add(1);
myCollection.Add(1);
The type of your array is int but you are trying to initialize it with an empty string. In C# you can declare an array in a few ways. You can declare it by providing a size as in:
int[] myArray = new int[size];
And initialize the values later.
An alternative is to instantly intialize it with values like this:
int[] myArray = new int[] { 1, 2, 3 };
Note that when using the second option, you shouldn't provied the size as the compiler will infere it.
I just saw this question and infact tried to answer it as well. But while answering I thought what could be the answer in case of C#? I am looking for some MSDN doc or any relevant source like in Java
What is the maximum dimension allowed for an array in C# like a[1][1][1][1]....[1]. I tried to search on SO but was not able to find.
The best I got up to was that "An array could theoretically have at most 2,147,483,647 elements, since it uses an int for indexing."
Also I am aware of the fact that Maximum Size that an Array can hold is
System.Int32.MaxValue
Kindly do let me know if this is a duplicate I will delete my question.
The limit is 32. I checked it with the code:
var b = Array.CreateInstance(typeof (int), new int[32]);
var a = Array.CreateInstance(typeof (int), new int[33]);
It applies for both 64 and 32-bit targeted applications.
Actually, I needn't have googled it. Straight from MSDN:
An element is a value in an Array. The length of an Array is the total number of elements it can contain. The rank of an Array is the number of dimensions in the Array. The lower bound of a dimension of an Array is the starting index of that dimension of the Array; a multidimensional Array can have different bounds for each dimension. An array can have a maximum of 32 dimensions.
About jagged arrays
However, I just noticed that the syntax you used in your edit (and also the syntax Peter used in the comments below) isn't how multi-dimensional arrays are defined in C#. Multi-dimensional arrays are defined as:
int[,,,] arr = new int[0,0,0];
And they are arrays of regular integers, that happen to have multiple dimensions. The syntax:
int[][][] = new int[0][][];
Defines something that in C# is called a jagged array. Jagged arrays are just regular, single-dimensional arrays that happen to have other arrays as elements. You can have multi-dimensional arrays of jagged arrays, or the other way around, as the MSDN article points out.
They are treated differently, however. Jagged arrays are just arrays of (effectiely) regular* objects, and as such, aren't regulated in the same way as multi-dimensional arrays. There is no theoretical limit to the number of "dimensions" a jagged array can have, and in fact the CLR happily let me run the following code:
Array arr = new int[0];
for (int i = 0; i < Int32.MaxValue - 1; i++) {
var newArr = Array.CreateInstance(arr.GetType(), 1);
newArr.SetValue(arr, 0);
arr = newArr;
}
Right up until my system became completely unresponsive and crashed. No exception was thrown.
Limiting the loop to 1000 iterations, you find that the type of the resulting array is:
System.Int32[][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][]...
In addition, although I can't find a source for this, the number of dimensions a jagged array can have probably isn't even limited by the single object memory limitation, because each array only has references to objects. For example, a jagged array of (eventually) Int32 with a nesting factor of, say, 1000 and 2 elements per array, would take up a negligible amount of memory (that is, on its own), but all the arrays combined would take up an exponential amount.
* With a little bit of syntax sugar. A more syntactically correct way of encoding jagged arrays would be: int[][] arr = new int[][0]; because int[] is the element type.
It's implementation defined
The number of dimensions an array can have appears to be implementation-defined. The number doesn't appear anywhere in the CLI specification, and the code provided by Peter below runs fine on Ideone's mono-2.8, where it appears the limit is 255.
Hi I was trying to find the number of elements in an array
byte[] salt = new byte[32];
now I only have mentioned size 32 so the Length Property of Array and Enumerable's Count Method will give me 32.
Even if I will iterate on this Array salt using for or foreach or any other looping construct it will iterate 32 times and on each index the value is 0 (i.e default value of byte)
Now suppose I do:
for (int i = 0; i < 5 ; i++)
{
salt[i] = 4-i;
}
And I want to know how many elements are inserted sequentially in Array starting from index 0, Here you may say are you fool you iterating it 5 times and you know the number is 5 , but I am having heavy looping and other logic (appending prepending to another arrays of byte) in it. *My question Is there any other inbuilt function that could give me this number 5 ? * Even if I iterate and check for first default value and break the loop there and get the count but there might be the chance last value inserted is 0 like above salt[4] is 0 so that iterating will give me the count 4 which is incorrect . If I am not wrong I think when we declare Array with size like 32 above 32 consecutive memory bytes are reserved and you are free to insert at any index from 0-31 and its your responsibility to keep the track where and how and how many elements are assigned to Array .I hope you got my point And thanks in advance for help.
An array is an array, and in .NET is initialized when it is allocated. Once it's initialized, the question of whether a given value is uninitialized or simply 0 isn't something that's possible to check. A 0 is a 0.
However, you can bypass that in several ways. You can use a List<int>, like #SLaks suggested, to have a dynamically allocated list that's only initialized with the elements you want.
You can also use, instead of an array of int, and array of int?, so a null value isn't the same as a 0.
Short answer is you can't, the array contains 32 integers, .net framework doesn't care if some of them are 0, so you can create your own function that counts how many integers from an array are different than 0, or keep a "count" when you assign values for array elements or something like that.
Or you can use another container, example a list and dynamically add or remove integers from it.
Ok, when you define an array as int[] myArray = int[32]; you are saying, I HAVE 32 ints. Not, create me space for 32 ints that I will fill in later. That's why count is giving you 32.
If you want something which you can genuinly add to and resize, you need to use a List (or one of it's relatives.
If you want to have a "cap" for a list, I found this :Maximum capacity collection in c#
i have an array like below
int[] array = new array[n];// n may be 2,3,4
example for N = 4
int[] array = new array[4];
array[0] = 2;
array[1] = 4;
array[2] = 6;
array[3] = 8;
how i calculate all unrepeated combination of this array without using linq may be within?
2,4,6,8
2,4,8,6
2,8,6,4
2,6,4,6
8,6,4,2
2,4,6,8
.......
.......
.......
Here's a pretty flexible C# implementation using iterators.
Well, given that you are looking for all unrepeated combinations, that means there will be N! such combinations... (so, in your case, N! = 4! = 24 such combinations).
As I'm in the middle of posting this, dommer has pointed out a good implementation.
Just be warned that it's is going to get really slow for large values of N (since there are N! permutations).
Think about the two possible states of the world to see if that sheds any light.
1) There are no dupes in my array(i.e. each number in the array is unique). In this case, how many possible permutations are there?
2) There is one single dupe in the array. So, of the number of permutations that you calculated in part one, how many are just duplicates
Hmmm, lets take a three element array for simplicity
1,3,5 has how many permutations?
1,3,5
1,5,3
3,1,5
3,5,1
5,1,3
5,3,1
So six permutations
Now what happens if we change the list to say 1,5,5?
We get
1,5,5
5,1,5
5,5,1
My question to you would be, how can you express this via factorials?
Perhaps try writing out all the permutations with a four element array and see if the light bulb goes off?
I am taking a C# class and I need help understanding the following code.
The code has an array which represents responses to a survey, with values 1 thru 10.
The output displays these ratings and the frequency of how many times a value was selected.
The following code is from my book, but I have modified it to just a basic example.
int[] responses = { 3, 2, 5, 6, 3, 5 , 4, 5, 5, 5};
int[] frequency = new int[7];
for (int answer = 0; answer < responses.Length; answer++)
++frequency[responses[answer]];
for (int rating = 1; rating < frequency.Length; rating++)
Console.WriteLine(rating + ", " + frequency[rating]);
Console.Read();
How does the line ++frequency[responses[answer]]; work? In looking at this, if I take reponses[answer] the first time through the loop, this would represent responses[0] which would be a 3, correct? This is where I get confused, what does the ++frequency part of this line do?
frequency[responses[answer]] = frequency[responses[answer]] + 1;
EDIT: I think it's pretty unclear to write it like that. As a personal preference, I don't like using unary operations (++x, x++, etc) on elements that have lots of indexes present.
It adds one to the frequency at that location in the array.
For example, the frequency at position 3 (from your example) will be increased by one after that line executes.
EDIT: So, in more detail, when answer = 0, responses[0] = 3, so frequency[3] gets one added to it.
The ++ could very easily be at the end of the command as well. In other words,
++frequency[responses[answer]];
is the same thing (IN THIS CASE) as using
frequency[responses[answer]]++;
Let's break it down: As you point out, on the first pass responses[answer] will evaluate to "3"
So this then looks like ++frequency[3]
The ++ is incrementing the value of the array at index 3 by 1
Simple enough?
I should also point out that applying the ++ before the array rather than after it does effect how the incrementing is executed (although it doesn't effect the results of this code).
For instance:
int n = 2;
int j = ++n;
int k = n++;
What are j and k?
j will be 3, and k will also be 3. This is because if you place the ++ before, it evaluates it first. If you place it at the end, it evaluates it after the rest of the expression.
If it helps, think of ++frequency[] as "frequency = frequency + 1".
If the ++ operator comes before the variable, then the increment is applied before the statement is executed. If the ++ comes afterwards, then the statement is executed and then the variable is incremented.
In this case, it doesn't matter, since incrementing before or after doesn't impact the logic.
Since "responses[answer]" evaluates to a number, that line of code is incrementing the frequency entry at that array index. So the first time through, answer is 0, so responses[answer] is 3, so the frequency[3] box is getting incremented by 1. The next time through, the frequency[2] box is incremented... etc. etc. etc.
frequency is an array, where all elements are initialized to 0 (the default value for an int). The line ++frequency[responses[answer]] will increment the frequency element pointed out by the integer found at responses[answer]. By putting the ++ in front of frequency, the array element will be incremented before the resulting value is returned.
You can read more about the ++ operator here.
In cases like this it's often useful to rewrite the code as you walk it.
When answer = 0
++frequency[responses[0]]
++frequency[3] since responses[0] = 3
frequency now looks like { 0, 0, 0, 1, 0, 0, 0 }
When answer = 1
++frequency[responses[1]]
++frequency[2] since responses[1] = 2
frequency now looks like { 0, 0, 1, 1, 0, 0, 0 }
And so on.
It means increment the value at frequency[ 3 ]. Where 3 is the return result from responses[answer]. Similarly the next iteration would increment the value at frequency[ 2 ].
The ++ operator in C# when applied to an integer will increment it by one.
The specific line you're looking at, frequency is an array of integers with 7 elements. Which is sort of confusing, because the way you explained it in your code, it would appear that this code would break with any value in the responses array above 6.
That issue aside, basically it's incrementing whichever index of the array it's accessing. So in your example responses[0] would be 3. So this line would find the value of frequency[3] and increment it by 1. Since integer arrays are initialized with all values at zero, then after the first iteration, frequency[3] would be 1. Then if there was another 3 later in your responses array, frequency[3] would be incremented again (i.e. responses[4]).
I hope this helps you.
The goal of the code snippet seems to be to determine the number of times each response appears in the 'responses' array. So, for your example set, frequency[3] should be 5, frequency[5] should be 5, etc.
So, the line you are asking about takes the current element from the responses array, and increments the associated value in the frequency array by 1, to indicate that the particular value has been observed in responses.
Once the entire code snippet executes, the frequency array contains the number of times each element from 0 to 7 was observed in the responses array.
It is using the frequency array to count how many times each response was entered. You could have a counter for each answer:
int numberOfOnes = 0;
int numberOfTwos = 0;
// Etc...
But that would be ugly programming and not as easy or efficient. Using the frequency array allows you do not use an if/else if block or a switch and makes your code easier to read.
Another thing about that frequency array.
int[] frequency = new int[7];
This initializes all the integers in the array to 0, that's why you can just start off by incrementing it instead of seeing if it was the first time for that specific response and then initializing it with 1 or something of that nature.
Good luck with all the fun C# you have ahead of you.