How does `*(*(p+1)+1)[7]` equal `p[1][8][0]`?

Question

char *p[2][3];

How does *(*(p+1)+1)[7] equal p[1][8][0]?

I thought *(*(p+1)+1)[7] would be the same as *(*(*(p+1)+1)+7) and that is equal to p[1][1][7], where am I wrong?

E: I don't understand why this is being downvoted...

Answer 1

Basic facts:

1. [] associates left to right
2. [] has higher precedence than the dereference operator *
3. a[i] is *(a+i)

So

  *(*(p+1)+1)[7] 
= *((*(p+1)+1)[7])   // rule 2
= *(*(*(p+1)+1+7))   // rule 3
= *(*(*(p+1)+8))    
= *(*(*(p+1)+8)+0)  
= p[1][8][0]         // rules 1 and 3

Answer 2

This depends on the relative precedence of [] vs unary * (the former is higher), and the essential identity that if p is (or evaluates to) a pointer and n is of integer type then p[n] means the same thing as *(p + n). You can therefore transform your starting expression according to the following steps, making repeated use, in both directions, of that identity:

1. *(*(p+1)+1)[7]
2. == *((*(p+1)+1)[7]) (The added parentheses express the operator precedence explicitly.)
3. == *((p[1]+1)[7])
4. == *(*((p[1]+1)+7))
5. == *(*(p[1]+1+7))
6. == *(*(p[1]+8))
7. == *(p[1][8])
8. == *(p[1][8] + 0)
9. == p[1][8][0]