In the following c/c code,

int main()
{
    int a[3] = {11, 22, 33};
    int *p[3];
    
    p[0] = &a[0];
    p[1] = &a[1];
    p[2] = &a[2];
    
    printf("(*p)[2] = %d\n",(*p)[2]);
    printf("*(p[2]) = %d\n",*(p[2]));
    return 0;
}

It turns out that (*p)[2] = *(p[2]) = a[2].

So, I understand that with int *p[3];, I am creating an array of three pointers, and with the subsequent three lines after int *p[3];, I am putting the address of a[0] into p[0], address of a[1] into p[1], and address of a[2] into p[2].

Then, with *(p[2]), I am retrieving the variable pointed by the address stored in p[2], which makes *(p[2]) = a[2]. I have two questions regarding this simple code:

1. how to understand that (*p)[2] also equals a[2]?

2. If p is no longer defined as an array of pointers to int variables, but an array of pointers to vectors of type int (std::vector in C ), does the equality (*p)[2] = *(p[2]) still hold?

CodePudding user response：

*p is the same as p[0], which is a pointer to a[0]. So (*p)[2] is a[2].

p[2] is a pointer to a[2], so *(a[2]) is also a[2].

Nothing magical going on here. It's just the way you have set up your pointers.

CodePudding user response：

It's because your array of pointers is an array of pointers that all point within the same array.

(*p)[2] effectively expands to (&a[0])[2] (because &a[0] is in p[0]), reading the second element of a by starting from &a[0] and skipping two elements.

*(p[2]) effectively expands to *(&a[2]) (because &a[2] is in p[2]), which is equivalent to (&a[2])[0], reading the second element of a by starting from &a[2] and not skipping any elements.

Either way, you end up reading the value from a[2], the only question is whether you're dereferencing a pointer that doesn't point to a[2] at an offset, or dereferencing a precalculated pointer to a[2] without an offset. You could get the same result with p[1][1] as well, it just doesn't look quite as symmetrical with the other two use cases.

Regardless, all of this works because the definition of p is weird; it's a toy to illustrate how pointers and indexing work.