Data commit

Task/Multi-dimensional-array/C/multi-dimensional-array-1.c

@@ -0,0 +1,37 @@
+/*Single dimensional array of integers*/
+int a[10];
+
+/*2-dimensional array, also called matrix of floating point numbers.
+This matrix has 3 rows and 2 columns.*/
+
+float b[3][2];
+
+/*3-dimensional array ( Cube ? Cuboid ? Lattice ?) of characters*/
+
+char c[4][5][6];
+
+/*4-dimensional array (Hypercube ?) of doubles*/
+
+double d[6][7][8][9];
+
+/*Note that the right most number in the [] is required, all the others may be omitted.
+Thus this is ok : */
+
+int e[][3];
+
+/*But this is not*/
+
+float f[5][4][];
+
+/*But why bother with all those numbers ? You can also write :*/
+int *g;
+
+/*And for a matrix*/
+float **h;
+
+/*or if you want to show off*/
+double **i[];
+
+/*you get the idea*/
+
+char **j[][5];

Task/Multi-dimensional-array/C/multi-dimensional-array-2.c

@@ -0,0 +1,27 @@
+#include<stdio.h>
+
+int main()
+{
+	int hyperCube[5][4][3][2];
+	
+	/*An element is set*/
+	
+	hyperCube[4][3][2][1] = 1;
+	
+	/*IMPORTANT : C ( and hence C++ and Java and everyone of the family ) arrays are zero based.
+	The above element is thus actually the last element of the hypercube.*/
+	
+	/*Now we print out that element*/
+	
+	printf("\n%d",hyperCube[4][3][2][1]);
+	
+	/*But that's not the only way to get at that element*/
+	printf("\n%d",*(*(*(*(hyperCube + 4) + 3) + 2) + 1));
+
+	/*Yes, I know, it's beautiful*/
+	*(*(*(*(hyperCube+3)+2)+1)) = 3;
+	
+	printf("\n%d",hyperCube[3][2][1][0]);
+	
+	return 0;
+}

Task/Multi-dimensional-array/C/multi-dimensional-array-3.c

@@ -0,0 +1,66 @@
+#include<stdlib.h>
+#include<stdio.h>
+
+/*The stdlib header file is required for the malloc and free functions*/
+
+int main()
+{
+	/*Declaring a four fold integer pointer, also called
+	a pointer to a pointer to a pointer to an integer pointer*/
+	
+	int**** hyperCube, i,j,k;
+
+	/*We will need i,j,k for the memory allocation*/
+	
+	/*First the five lines*/
+	
+	hyperCube = (int****)malloc(5*sizeof(int***));
+	
+	/*Now the four planes*/
+	
+	for(i=0;i<5;i++){
+		hyperCube[i] = (int***)malloc(4*sizeof(int**));
+		
+		/*Now the 3 cubes*/
+		
+		for(j=0;j<4;j++){
+			hyperCube[i][j] = (int**)malloc(3*sizeof(int*));
+			
+			/*Now the 2 hypercubes (?)*/
+			
+			for(k=0;k<3;k++){
+				hyperCube[i][j][k] = (int*)malloc(2*sizeof(int));
+			}
+		}
+	}
+	
+	/*All that looping and function calls may seem futile now,
+	but imagine real applications when the dimensions of the dataset are
+	not known beforehand*/
+	
+	/*Yes, I just copied the rest from the first program*/
+	
+	hyperCube[4][3][2][1] = 1;
+	
+	/*IMPORTANT : C ( and hence C++ and Java and everyone of the family ) arrays are zero based.
+	The above element is thus actually the last element of the hypercube.*/
+	
+	/*Now we print out that element*/
+	
+	printf("\n%d",hyperCube[4][3][2][1]);
+	
+	/*But that's not the only way to get at that element*/
+	printf("\n%d",*(*(*(*(hyperCube + 4) + 3) + 2) + 1));
+
+	/*Yes, I know, it's beautiful*/
+	*(*(*(*(hyperCube+3)+2)+1)) = 3;
+	
+	printf("\n%d",hyperCube[3][2][1][0]);
+	
+	/*Always nice to clean up after you, yes memory is cheap, but C is 45+ years old,
+	and anyways, imagine you are dealing with terabytes of data, or more...*/
+	
+	free(hyperCube);
+	
+	return 0;
+}