Converting multidimensional arrays to pointer notation

Question

For the following value-of an array, I can do:

array[i][j][k] = *(*(*(array+i)+j)+k);
---------------------------------

int xar[3] = {1,2,3};
printf("%d | %d\n", xar[1], *(xar+1));

int xar2[2][3] = {{1,2,3}, {4,5,6}};
printf("%d | %d\n", xar2[1][2], *(*(xar2+1)+2));

int xar3[2][2][2] = {
                      {{1, 2},{3, 4}}, 
                      {{5, 6}, {7, 8}}
                    };
printf("%d | %d\n", xar3[1][1][1], *(*(*(xar3+1)+1)+1));

2 | 2
6 | 6
8 | 8

However, I'm having a tougher time trying to get the address of an array element and why that works, for example, if I have:

&array[i][j][k];

array+i gives me the equivalent of &array[i], but then how would I grab the full inner item?

@ChrisDodd ok but could you please explain how that would work? — David542, Feb 06 '21 at 09:45
Does this answer your question? [Pointer address in a C multidimensional array](https://stackoverflow.com/questions/2003745/pointer-address-in-a-c-multidimensional-array) — Rane, Feb 06 '21 at 09:50
Yes, it sounds correct now, but please note that there may be some dupes around. — Bob__, Feb 06 '21 at 09:57

score 2 · Answer 1 · answered Feb 06 '21 at 09:49

2

Note that &*array will cancel each-other out, so we can do:

 array[i][j][k] =  *(*(*(array+i)+j)+k);
&array[i][j][k] = &*(*(*(array+i)+j)+k);   // &* can be eliminated
&array[i][j][k] =    *(*(array+i)+j)+k;

answered Feb 06 '21 at 09:49

David542

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Converting multidimensional arrays to pointer notation

1 Answers1