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Ingy döt Net 2025-02-27 18:35:13 -05:00
parent 8e4e15fa56
commit 72eb4943cb
1853 changed files with 35514 additions and 9441 deletions
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37
Task/Multi-dimensional-array/C/multi-dimensional-array-1.c
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/*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];

27
Task/Multi-dimensional-array/C/multi-dimensional-array-2.c
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#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;
}

66
Task/Multi-dimensional-array/C/multi-dimensional-array-3.c
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#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;
}

31
Task/Multi-dimensional-array/C/multi-dimensional-array.c Normal file
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/* an array of ten ints */
int a[10];
/* a 2D-array of floats with three rows and two columns */
float b[3][2];
/*
these would be ordered in memory as
b[0][0] b[0][1] b[1][0] b[1][1] b[2][0] b[2][1]
for example:
*/
b[0][0] = 1.0;
b[0][1] = 2.0;
b[1][0] = 3.0;
b[1][1] = 4.0;
b[2][0] = 5.0;
b[2][1] = 6.0;
/*
now these would be stored in memory as:
+----+----+----+----+----+----+
| 1.0| 2.0| 3.0| 4.0| 5.0| 6.0|
+----+----+----+----+----+----+
*/
/* a 3D-array of chars */
char c[4][5][6];
/* etc. */