2D-array kan defineres som en rekke arrays. 2D-matrisen er organisert som matriser som kan representeres som samlingen av rader og kolonner.
tilfeldig i c
Imidlertid er 2D-matriser opprettet for å implementere en relasjonsdatabase-liknende datastruktur. Det gjør det enkelt å holde store mengder data på en gang som kan sendes til et hvilket som helst antall funksjoner der det er nødvendig.
Hvordan erklære 2D Array
Syntaksen for å deklarere todimensjonal matrise er veldig lik den for en endimensjonal matrise, gitt som følger.
int arr[max_rows][max_columns];
Det produserer imidlertid datastrukturen som ser ut som følgende.
Bildet ovenfor viser den todimensjonale matrisen, elementene er organisert i form av rader og kolonner. Første element i den første raden er representert av a[0][0] der tallet vist i den første indeksen er nummeret på den raden mens tallet som vises i den andre indeksen er nummeret på kolonnen.
Hvordan får vi tilgang til data i en 2D-array
På grunn av det faktum at elementene i 2D-matriser kan fås tilfeldig. I likhet med endimensjonale arrays, kan vi få tilgang til de individuelle cellene i en 2D-array ved å bruke indeksene til cellene. Det er to indekser knyttet til en bestemt celle, den ene er radnummeret mens den andre er kolonnenummeret.
Imidlertid kan vi lagre verdien som er lagret i en bestemt celle i en 2D-matrise til en variabel x ved å bruke følgende syntaks.
int x = a[i][j];
hvor i og j er henholdsvis rad- og kolonnenummeret til cellen.
Vi kan tilordne hver celle i en 2D-matrise til 0 ved å bruke følgende kode:
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> Antall elementer som kan være til stede i en 2D-matrise vil alltid være lik ( antall rader * antall kolonner ).
Eksempel: Lagre brukerens data i en 2D-array og skrive den ut.
C Eksempel:
#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++)> hvor B.A. er basisadressen eller adressen til det første elementet i matrisen a[0][0] .
Eksempel:
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
Etter kolonne hovedordre
Hvis matrise er deklarert av a[m][n] der m er antall rader mens n er antall kolonner, beregnes adressen til et element a[i][j] i matrisen lagret i rad hovedrekkefølge som ,
Address(a[i][j]) = ((j*m)+i)*Size + BA
der BA er basisadressen til matrisen.
Eksempel:
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