25.MultiIndexed Arrays With Pointers

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25.MultiIndexedArraysWithPointers

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deepak4u106

I want to make the following pattern using 2-INDEX array

1

12

123

1234

Can this code help to make the above pattern?

#include<stdio.h>

#include<stdlib.h>

int main(){

int r,c,i,j;

printf("Enter the value of row and col\n");

scanf("%d%d", &r,&c);

int *arr;

arr = (int*)malloc(r*c*sizeof(int));

for(i = 0 ; i < r ; ++i){

for(j = 0 ; j < c ; ++j){

scanf("%d", ((arr + i) + j));

}

for(i = 0 ; i < r ; ++i){

for(j = 0 ; j < c ; ++j){

printf("%d", *((arr + i) + j));

}

printf("\n");

}

ritde1

No

jagjeet.singh9650

Change the check condition in second nested loop , in the nested loop change j<c to j<=r, you will get your desired pattern. Change only where you are printing the values to obtain desired pattern.

jyotipandey9813

hi deepak

here is the solution of your problem

you made 2 mistakes

1. first mistake is : second nested loop will go till j<=i instead of j<c.

2. and second biggest mistake is , how you store the elements in 2 index array .

it should be : scanf("%d", ((arr + i*c) + j));

printf("%d", *((arr + i*c) + j));

you are going good ,just need to focus on memory representation of 2d index

deepak4u106

When ever we change the value of column .....pattern going to change or not??

deepak4u106

And thank you Jagjeet and Jyoti i will try.

jyotipandey9813

yes , pattern will change after changing the value of row or/and column.

swatiupendra1999

Hi am stuck at a pattern question :

----*****

---****

--***

-**

*

-**

--***

---****

----*****

I have issue with the lower highlighted star area.

swatiupendra1999

#include<stdio.h>

#include<stdlib.h>

int main()

{

int *arr,i,l,j,s,r,c,k;

printf("%s:BEGIN\n",__func__);

printf("Enter the value of Rows\n");

scanf("%d",&r);

printf("Enter the value of Coloumn\n");

scanf("%d",&c);

arr = (int*)malloc(sizeof(r*c));

if(!arr)

{

perror("malloc");

exit(EXIT_FAILURE);

}

swatiupendra1999

for(i=0;i<r/2;i++)

{

for(j=0;j<c/2-i;j++)

{

*(arr + i*c+ j)='-';

}

for(j=c/2;j<c-i;j++)

{

*(arr + i*c+ j)='*';

}

k=r/2;

l=1;

printf("%d\n",r/2);

for(i=r/2;i<r;i++)

{

for(j=0 ; j<r/2-k ; j++)

{

*(arr+ i*c + j)='-';

}

/* for(j=0;j<=c/2-k;j+=l)

{

*(arr + i*c+ j)='*';

}

*/

k--;

// l++;

}

printf("%s:PATTERN BEGIN\n",__func__);

for(i=0;i<r;i++)

{

for(j=0;j<c;j++)

{

printf("%c",*(arr + c*i + j ));

}

printf("\n");

}

printf("%s:PATTERN END\n",__func__);

printf("%s:END\n",__func__);

return 0;

}

ABOVE IS THE CODE I WROTE

swatiupendra1999

----*****

---****

--***

-**

-

--

---

----

deepak4u106

In second nested loop where row lies between "r/2" to "r"

the second condition should increase by 1 in first iteration and 2 in second iteration and 3 in third iteration.

In 2-D arrays we are assigning the value at positions ie 2:2,2:3,2:4 , not in the indices.

Below is the program would help you to better understand about above explanation

deepak4u106

#include <stdio.h>

#include <string.h>

int main()

{

char *ptr;

int r,c,space,star,rowIndex,j = 0,k = 0;

printf("%s : Begin\n",__func__);

printf("Enter the Rows\n");

scanf("%d", &r);

printf("Enter the Col\n");

scanf("%d", &c);

ptr = (char*)malloc(sizeof(char)*r*c);

for(rowIndex = 0 ; rowIndex < r/2 ; rowIndex++)

{

for(space = 0 ; space < (c/2)-rowIndex ; space++)

{

*(ptr + rowIndex*c + space) = '-';

}

for(star = c/2 ; star < c-rowIndex ; star++)

{

*(ptr + rowIndex*c + star) = '*';

}

j = 1;

for(rowIndex = r/2 ; rowIndex < r ; rowIndex++)

{

for(space = 0 ; space < k ; space++)

{

*(ptr + rowIndex*c + space) = '-';

}

for(star = k ; star < k + j ; star++)

{

*(ptr + rowIndex*c + star) = '*';

}

j++;

k++;

}

for(rowIndex = 0 ; rowIndex < r ; rowIndex++)

{

for(j = 0 ; j < c ; j++)

{

printf("%c", *(ptr + rowIndex*c + j));

}

printf("\n");

}

printf("%s : End\n",__func__);

return 0;

}

swatiupendra1999

thank you

swatiupendra1999

How can we fill every position in an 2 indexed array with null .

ayushdhiman117

use memsetfunction. syntax of memset function--->memset(pointername,value,size);

ayushdhiman117

example -->

char *jrr;

jrr=(char*)malloc(sizeof (char)* row*col);

memset(jrr,'\0',row*col);

ayushdhiman117

please show the meaning of null pointer.

example-->

int *krr;

krr=(int*)malloc(sizeof(int)*size);

if(krr==NULL)

{

perror("malloc");

exit(EXIT_FAILURE);

}

if(krr==NULL) //condition is true when i give size = -1.

compiler are show the result---> malloc are not allocated.

kya ye he hoota hai NULL ka meaning.

deepak4u106

@ayushdhiman117

size should be in positive integer number else it will throw an error of not allocating the memory

deepak4u106

swatiupendra1999

#include <string.h>

memset(char pointer name, The value to fill in the memory or block of char pointer , size of char pointer);

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

int main()

{

int row = 4;

int col = 5;

char *s = (char*)malloc(sizeof(char) * row * col);

//Now filling the whole places or blocks or bytes with '0'

memset( s , '\0' , row * col); //Pre defined function in library <string.h>

/*

This is what we have done in string it reset the whole garbage value with NULL character you can fill with any character ie space , star, any, or 0 - 9

\0 \0 \0 \0

*/

}

25.MultiIndexed Arrays With Pointers

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