Ein 2D-Array kann als Array von Arrays definiert werden. Das 2D-Array ist in Form von Matrizen organisiert, die als Sammlung von Zeilen und Spalten dargestellt werden können.
Java-Ganzzahl
Allerdings werden 2D-Arrays erstellt, um eine relationale, datenbankähnliche Datenstruktur zu implementieren. Es ermöglicht die einfache Speicherung großer Datenmengen auf einmal, die bei Bedarf an eine beliebige Anzahl von Funktionen übergeben werden können.
So deklarieren Sie ein 2D-Array
Die Syntax der Deklaration eines zweidimensionalen Arrays ist der eines eindimensionalen Arrays sehr ähnlich, wie folgt.
int arr[max_rows][max_columns];
Es erzeugt jedoch die Datenstruktur, die wie folgt aussieht.
Das obige Bild zeigt das zweidimensionale Array, die Elemente sind in Form von Zeilen und Spalten organisiert. Das erste Element der ersten Zeile wird durch a[0][0] dargestellt, wobei die im ersten Index angezeigte Zahl die Nummer dieser Zeile ist, während die im zweiten Index angezeigte Zahl die Nummer der Spalte ist.
Wie greifen wir auf Daten in einem 2D-Array zu?
Aufgrund der Tatsache, dass auf die Elemente von 2D-Arrays wahlfrei zugegriffen werden kann. Ähnlich wie bei eindimensionalen Arrays können wir auf die einzelnen Zellen in einem 2D-Array zugreifen, indem wir die Indizes der Zellen verwenden. Einer bestimmten Zelle sind zwei Indizes zugeordnet: einer ist die Zeilennummer und der andere die Spaltennummer.
Wir können jedoch den in einer bestimmten Zelle eines 2D-Arrays gespeicherten Wert mithilfe der folgenden Syntax in einer Variablen x speichern.
int x = a[i][j];
wobei i und j die Zeilen- bzw. Spaltennummer der Zelle sind.
Wir können jeder Zelle eines 2D-Arrays den Wert 0 zuweisen, indem wir den folgenden Code verwenden:
for ( int i=0; i<n ;i++) { for (int j="0;" j<n; j++) a[i][j]="0;" } < pre> <h2>Initializing 2D Arrays </h2> <p>We know that, when we declare and initialize one dimensional array in C programming simultaneously, we don't need to specify the size of the array. However this will not work with 2D arrays. We will have to define at least the second dimension of the array. </p> <p>The syntax to declare and initialize the 2D array is given as follows. </p> <pre> int arr[2][2] = {0,1,2,3}; </pre> <p>The number of elements that can be present in a 2D array will always be equal to ( <strong>number of rows * number of columns</strong> ). </p> <p> <strong>Example :</strong> Storing User's data into a 2D array and printing it. </p> <p> <strong>C Example : </strong> </p> <pre> #include void main () { int arr[3][3],i,j; for (i=0;i<3;i++) { for (j="0;j<3;j++)" printf('enter a[%d][%d]: ',i,j); scanf('%d',&arr[i][j]); } printf('
printing the elements ....
'); for(i="0;i<3;i++)" printf('
'); printf('%d ',arr[i][j]); < pre> <h3>Java Example</h3> <pre> import java.util.Scanner; publicclass TwoDArray { publicstaticvoid main(String[] args) { int[][] arr = newint[3][3]; Scanner sc = new Scanner(System.in); for (inti =0;i<3;i++) { for(intj="0;j<3;j++)" system.out.print('enter element'); arr[i][j]="sc.nextInt();" system.out.println(); } system.out.println('printing elements...'); for(inti="0;i<3;i++)" system.out.print(arr[i][j]+' '); < pre> <h3>C# Example </h3> <pre> using System; public class Program { public static void Main() { int[,] arr = new int[3,3]; for (int i=0;i<3;i++) { for (int j="0;j<3;j++)" console.writeline('enter element'); arr[i,j]="Convert.ToInt32(Console.ReadLine());" } console.writeline('printing elements...'); i="0;i<3;i++)" console.writeline(); console.write(arr[i,j]+' '); < pre> <h2>Mapping 2D array to 1D array </h2> <p>When it comes to map a 2 dimensional array, most of us might think that why this mapping is required. However, 2 D arrays exists from the user point of view. 2D arrays are created to implement a relational database table lookalike data structure, in computer memory, the storage technique for 2D array is similar to that of an one dimensional array. </p> <p>The size of a two dimensional array is equal to the multiplication of number of rows and the number of columns present in the array. We do need to map two dimensional array to the one dimensional array in order to store them in the memory.</p> <p>A 3 X 3 two dimensional array is shown in the following image. However, this array needs to be mapped to a one dimensional array in order to store it into the memory. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-2.webp" alt="DS 2D Array"> <br> <p>There are two main techniques of storing 2D array elements into memory </p> <h3>1. Row Major ordering </h3> <p>In row major ordering, all the rows of the 2D array are stored into the memory contiguously. Considering the array shown in the above image, its memory allocation according to row major order is shown as follows. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-3.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> row of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last row.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-4.webp" alt="DS 2D Array"> <br> <h3>2. Column Major ordering </h3> <p>According to the column major ordering, all the columns of the 2D array are stored into the memory contiguously. The memory allocation of the array which is shown in in the above image is given as follows.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-5.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> column of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last column of the array. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-6.webp" alt="DS 2D Array"> <br> <h2>Calculating the Address of the random element of a 2D array </h2> <p>Due to the fact that, there are two different techniques of storing the two dimensional array into the memory, there are two different formulas to calculate the address of a random element of the 2D array. </p> <h3>By Row Major Order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = B. A. + (i * n + j) * size </pre> <p>where, B. A. is the base address or the address of the first element of the array a[0][0] . </p> <p> <strong>Example : </strong> </p> <pre> a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer </pre> <h3>By Column major order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = ((j*m)+i)*Size + BA </pre> <p>where BA is the base address of the array. </p> <p> <strong>Example:</strong> </p> <pre> A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes </pre> <hr></3;i++)></pre></3;i++)></pre></3;i++)></pre></n> Die Anzahl der Elemente, die in einem 2D-Array vorhanden sein können, ist immer gleich ( Anzahl der Zeilen * Anzahl der Spalten ).
Beispiel : Speichern der Benutzerdaten in einem 2D-Array und Drucken.
C-Beispiel:
#include void main () { int arr[3][3],i,j; for (i=0;i<3;i++) { for (j="0;j<3;j++)" printf(\'enter a[%d][%d]: \',i,j); scanf(\'%d\',&arr[i][j]); } printf(\'
printing the elements ....
\'); for(i="0;i<3;i++)" printf(\'
\'); printf(\'%d \',arr[i][j]); < pre> <h3>Java Example</h3> <pre> import java.util.Scanner; publicclass TwoDArray { publicstaticvoid main(String[] args) { int[][] arr = newint[3][3]; Scanner sc = new Scanner(System.in); for (inti =0;i<3;i++) { for(intj="0;j<3;j++)" system.out.print(\'enter element\'); arr[i][j]="sc.nextInt();" system.out.println(); } system.out.println(\'printing elements...\'); for(inti="0;i<3;i++)" system.out.print(arr[i][j]+\' \'); < pre> <h3>C# Example </h3> <pre> using System; public class Program { public static void Main() { int[,] arr = new int[3,3]; for (int i=0;i<3;i++) { for (int j="0;j<3;j++)" console.writeline(\'enter element\'); arr[i,j]="Convert.ToInt32(Console.ReadLine());" } console.writeline(\'printing elements...\'); i="0;i<3;i++)" console.writeline(); console.write(arr[i,j]+\' \'); < pre> <h2>Mapping 2D array to 1D array </h2> <p>When it comes to map a 2 dimensional array, most of us might think that why this mapping is required. However, 2 D arrays exists from the user point of view. 2D arrays are created to implement a relational database table lookalike data structure, in computer memory, the storage technique for 2D array is similar to that of an one dimensional array. </p> <p>The size of a two dimensional array is equal to the multiplication of number of rows and the number of columns present in the array. We do need to map two dimensional array to the one dimensional array in order to store them in the memory.</p> <p>A 3 X 3 two dimensional array is shown in the following image. However, this array needs to be mapped to a one dimensional array in order to store it into the memory. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-2.webp" alt="DS 2D Array"> <br> <p>There are two main techniques of storing 2D array elements into memory </p> <h3>1. Row Major ordering </h3> <p>In row major ordering, all the rows of the 2D array are stored into the memory contiguously. Considering the array shown in the above image, its memory allocation according to row major order is shown as follows. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-3.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> row of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last row.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-4.webp" alt="DS 2D Array"> <br> <h3>2. Column Major ordering </h3> <p>According to the column major ordering, all the columns of the 2D array are stored into the memory contiguously. The memory allocation of the array which is shown in in the above image is given as follows.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-5.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> column of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last column of the array. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-6.webp" alt="DS 2D Array"> <br> <h2>Calculating the Address of the random element of a 2D array </h2> <p>Due to the fact that, there are two different techniques of storing the two dimensional array into the memory, there are two different formulas to calculate the address of a random element of the 2D array. </p> <h3>By Row Major Order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = B. A. + (i * n + j) * size </pre> <p>where, B. A. is the base address or the address of the first element of the array a[0][0] . </p> <p> <strong>Example : </strong> </p> <pre> a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer </pre> <h3>By Column major order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = ((j*m)+i)*Size + BA </pre> <p>where BA is the base address of the array. </p> <p> <strong>Example:</strong> </p> <pre> A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes </pre> <hr></3;i++)></pre></3;i++)></pre></3;i++)> Dabei ist B. A. die Basisadresse oder die Adresse des ersten Elements des Arrays a[0][0] .
Beispiel :
a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer
Nach Hauptreihenfolge der Spalte
Wenn das Array durch a[m][n] deklariert wird, wobei m die Anzahl der Zeilen und n die Anzahl der Spalten ist, wird die Adresse eines Elements a[i][j] des in Zeilenhauptreihenfolge gespeicherten Arrays wie folgt berechnet ,
Address(a[i][j]) = ((j*m)+i)*Size + BA
Dabei ist BA die Basisadresse des Arrays.
Beispiel:
A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes