Writing array reference using pointer arithmetic

Question

let consider the following fragment of Code:

#include<stdio.h>
main()
{
 int count[100][10];
 *(count + (44*10)+8)=99;
 printf("%d",count[44][8]);
}

What is the wrong with it?

possible duplicate of [Why does int\*\[\] decay into int\*\* but not int\[\]\[\]?](http://stackoverflow.com/questions/14183546/why-does-int-decay-into-int-but-not-int) — ecatmur, Jan 29 '15 at 11:32

Gopi · Answer 1 · 2015-01-29T11:35:02.773

2

count[44][8]

is not initialized and you are trying to print the value of it which is UB.

a[i][j] = *(a[i] + j); 
a[i][j] = *(*(a+i) + j);

So if you want to initialize count[44][8] then do

*(count[44] + 8) = 10; /* or *(*(count + 44) + 8) = 10 */
printf("%d",count[44][8]);

edited Jan 29 '15 at 11:35

answered Jan 29 '15 at 11:29

Gopi

Would U please explain ur code...! is it a rule a[i][j] = *(a[i] + j); a[i][j] = *(*(a+i) + j) for accessing 2d array @Gopi – Black Swan Jan 29 '15 at 12:16
int count[100][10],*countP; countP=count; *(countP[44]+8)=99; it is not working for this fragment. – Black Swan Jan 29 '15 at 12:27
@BlackSwan - that is the rule. Any array is simply a block of addresses in memory. You can address any specific address with either normal array notation `count[44][8]` or by specifying the `offset` in memory from the base address of count (e.g. `*(*(count + 44) + 8)`). The basic equivalence is easier understood for a 1D array: `a[i] = *(a + i)`. For a 2D array, you simply have an additional layer: `a[i][j] = *(*(a + i) + j)` – David C. Rankin Mar 27 '15 at 06:08

score 2 · Answer 2 · edited May 23 '17 at 12:34

2

Array-to-pointer decay only works for one level; so int count[100][10]; decays to int (*)[100] (Why does int*[] decay into int** but not int[][]?).

You can either cast count to int* or use &count[0][0] to get an int* pointer to the first element of the 2D array.

edited May 23 '17 at 12:34

Community

answered Jan 29 '15 at 11:31

ecatmur

i am not familiar c++, only with C. it worked in ur way, but i am still confused that why i have to cast it...! – Black Swan Jan 29 '15 at 12:02
@BlackSwan see Oliver Charlesworth's answer: http://stackoverflow.com/a/14183554/567292 - array-to-pointer decay works exactly the same in C and C++, it's just that C++ gives more ways to observe it. – ecatmur Jan 29 '15 at 12:04

score 1 · Answer 3 · answered Jan 29 '15 at 11:28

1

*(count + (44*10)+8)=99; should be

*(count[0] + (44*10)+8)=99;

Type of countp[0] can be reinterpreted as int * as you want.

Type of count is int [100][10] so adding some big number to it would go 10 times ahead as you want and access to that location would lead to UB.

Anopter way to write the same is:

*( *(count + 44) + 8 )=99;

answered Jan 29 '15 at 11:28

Mohit Jain

1

An explanation would be nice. – Ed Heal Jan 29 '15 at 11:29
@EdHeal I accidentally hit some key + enter key which submitted answer before completion. I accept (+1 for reminding) it was incomplete. – Mohit Jain Jan 29 '15 at 11:30
is ur answer complete now? @Mohit Jain – Black Swan Jan 29 '15 at 11:54
@BlackSwan Yes, it complete now. – Mohit Jain Jan 29 '15 at 12:22

3 Answers3