I have a question about a pointer to 2d array. If an array is something like
int a[2][3]; then, is this a pointer to array a?
int (*p)[3] = a; If this is correct, I am wondering what does [3] mean from int(*p)[3]?
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6 Answers
int a[2][3]; a is read as an array 2 of array 3 of int which is simply an array of arrays. When you write,
int (*p)[3] = a;
It declares p as a pointer to the first element which is an array. So, p points to the array of 3 ints which is a element of array of arrays.
Consider this example:
int a[2][3] +----+----+----+----+----+----+ | | | | | | | +----+----+----+----+----+----+ \_____________/ | | | p int (*p)[3] Here, p is your pointer which points to the array of 3 ints which is an element of array of arrays.
1 Comment
Matthew
wouldn't "p" become an array of 3 integer pointers, with the first value pointing to the first value of "a"?
Rather than referring to int[2][3] as a '2d array', you should consider it to be an 'array of arrays'. It is an array with two items in it, where each item is itself an array with 3 ints in it.
int (*p)[3] = a; You can use p to point to either of the two items in a. p points to a three-int array--namely, the first such item. p+1 would point to the second three-int array. To initialize p to point to the second element, use:
int (*p)[3] = &(a[1]); The following are equivalent ways to point to the first of the two items.
int (*p)[3] = a; // as before int (*p)[3] = &(a[0]); 
Stricly speaking, no, int (*p)[3] = a; is not a pointer to a. It is a pointer to the first element of a. The first element of a is an array of three ints. p is a pointer to an array of three ints.
A pointer to the array a would be declared thus:
int (*q)[2][3] = &a; The numeric value of p and q are likely (or maybe even required to be) the same, but they are of different types. This will come into play when you perform arithmetic on p or q. p+1 points to the second element of array a, while q+1 points to the memory just beyond the end of array a.
Remember: cdecl is your friend: int a[2][3], int (*q)[2][3].
Comments
The [3] is a part of the type. In this case p is a pointer to an array of size 3 which holds ints.
The particular type of an array always includes its size, so that you have the types int *[3] or int *[5], but not just int *[] which has undefined size.
int *x[20]; /* type of x is int *[20], not just int *[] */ int y[10][10]; /* type of y is int[10][10], not just int[][] */ answered Dec 23, 2011 at 15:06
Blagovest Buyukliev
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2 Comments
user1047092
Thanks for your comment. When you say size, does it mean (*p) has [0],[1]],[2] ? If this is the case, where is a room for [2]? Or does it points to [0][0] and [0][1]?
Blagovest Buyukliev
p is simply a pointer to an array of 3 ints. It doesn't know that the first dimension of a is 2, because it is assigned to the first element of a, which is exactly of type int[3].you can point to 2d array like 1d array
#include <iostream> int main() { int array[2][2] = {{0,1}, {2,3}}; // array int *ptr; ptr=(int*)array; std::cout << *(ptr) << '\n';//out 0 std::cout << *(ptr+1) << '\n';//out 1 std::cout << *(ptr+2) << '\n';//out 2 std::cout << *(ptr+3) << '\n';//out 3 } 
Comments
Also note:
int *p[5] // p is an array of 5 pointers int (*p)[5] // p points to an array of 5 ints int (*(p+5))[10] // p is a pointer to a structure where the structure's 5th element has 10 ints . 
