1

I want to skip p from the start of {0,1,2,3}, to {4,5,6,7} etc. How do I do that?

#include <stdio.h>

int main() {

    char td[6][4] = { 
                      {0, 1, 2, 3}, 
                      {4, 5, 6, 7}, 
                      {8, 9, 10, 11}, 
                      {12, 13, 14, 15}, 
                      {16, 17, 18, 19}, 
                      {20, 21, 22, 23}  
                    };
    char* p = *td;  

    printf("Address of td: \t%p, value=%u\n", td, (int)**td);
    printf("Address of p: \t%p, value=%u\n", p, (int)*p);
    p++;   
    /* How do I skip to the start of {4,5,6,7} (ie to be pointing at 4) ? */ 
    printf("Address of p: \t%p, value=%u\n", p, (int)*p);

    return 0;
}
Grijesh Chauhan
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Angus Comber
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3 Answers3

1

You should first read this answer: Declaration-2: Where I explained how char 2-D array stored in memory.

char* p points to td[0][0] element in array that is 0. If you wants to increment p such that p point to 4 that is td[1][0] then increment p by number of cols = 4 (array is row = 6 * col = 4).

p = p + 4;

Check this working code. I applied concept of row-major.

Community
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Grijesh Chauhan
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0

In order to skip to the next row, you need a pointer which points to the row containing four elements not a pointer to a single element.

So, the pointer would look like char (*p)[4] = td now when you increment p by one it will be pointing to the next row ({4,5,6,7})

Also, your initialization of the character pointer is incorrect, it should be char *p = td

Uchia Itachi
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0

td + 1 will also skip by 4 elements. Because td is pointing to first array consist of 4 elements.It is a pointer to a array

printf("Address of p: \t%p, value=%u\n", *(td+1), **(td+1));

example

char a[row][col];

for(i = 0 ; i < row*col ; i++)
  printf("%c", *(*a+i) );

this will print the whole 2d array with just one loop.

Krv Perera
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