La matriu 2D es pot definir com una matriu de matrius. La matriu 2D s'organitza com a matrius que es poden representar com la col·lecció de files i columnes.
Tanmateix, es creen matrius 2D per implementar una estructura de dades semblant a una base de dades relacional. Proporciona facilitat per mantenir una gran quantitat de dades alhora que es poden passar a qualsevol nombre de funcions on sigui necessari.
Com declarar la matriu 2D
La sintaxi de declarar una matriu bidimensional és molt semblant a la d'una matriu unidimensional, donada de la següent manera.
base de dades java jdbc
int arr[max_rows][max_columns];
tanmateix, produeix l'estructura de dades que sembla la següent.
La imatge de dalt mostra la matriu bidimensional, els elements estan organitzats en forma de files i columnes. El primer element de la primera fila es representa amb a[0][0] on el número que es mostra al primer índex és el número d'aquesta fila mentre que el número que es mostra al segon índex és el número de la columna.
Com accedim a les dades en una matriu 2D
A causa del fet que es pot accedir aleatòriament als elements de les matrius 2D. De manera similar a les matrius unidimensionals, podem accedir a les cel·les individuals en una matriu 2D mitjançant els índexs de les cel·les. Hi ha dos índexs units a una cel·la concreta, un és el seu número de fila i l'altre és el seu número de columna.
Tanmateix, podem emmagatzemar el valor emmagatzemat en qualsevol cel·la particular d'una matriu 2D en alguna variable x utilitzant la sintaxi següent.
int x = a[i][j];
on i i j són el número de fila i columna de la cel·la respectivament.
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Podem assignar cada cel·la d'una matriu 2D a 0 mitjançant el codi següent:
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> El nombre d'elements que poden estar presents en una matriu 2D sempre serà igual a ( nombre de files * nombre de columnes ).
Exemple: Emmagatzemar les dades de l'usuari en una matriu 2D i imprimir-les.
C Exemple:
#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++)> on, B. A. és l'adreça base o l'adreça del primer element de la matriu a[0][0] .
Exemple:
proves de rendiment
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
Per ordre major de la columna
Si la matriu es declara per a[m][n] on m és el nombre de files mentre que n és el nombre de columnes, llavors l'adreça d'un element a[i][j] de la matriu emmagatzemada en l'ordre principal de les files es calcula com ,
Address(a[i][j]) = ((j*m)+i)*Size + BA
on BA és l'adreça base de la matriu.
Exemple:
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