在这种情况下，C 中的 int** 是什么意思？

Question

Here is the context:

int *t[10];
int n;

I am being told that tn is of the type int**. I don't exactly get what int** means, is it a pointer of a pointer? Why would the subtraction of a table of pointers - int would give a pointer of pointer of an int? When we refer to *t[0] do we refer to int* p to the pointer itself because it is an element of the table or do we implicitly need a pointer to point at the slot than have the pointer point to the another place?

Thanks in advance for explaining this to me.

Answer 1

I don't exactly get what int** means, is it a pointer of a pointer?

Yes.

• int = integer
• int * = pointer-to-integer
• int ** = pointer-to-(pointer-to-integer)
• int *** = pointer-to-(pointer-to-(pointer-to-integer))
• (and so on)

Why would the subtraction of a table of pointers - int would give a pointer of pointer of an int?

Because in C (and C++), an array decays into a pointer to the first item when necessary. For example, int *t[10] is an array of 10 pointer-to-int items. t can decay into a pointer to t[0] , ie a pointer-to-(pointer-to-int), int ** . That pointer can then be used for pointer arithmetic (like subtraction).

So, subtracting n from t would give you an int ** that is pointing n items "before" the beginning of your 10-item array (which BTW would not be a safe pointer to use, unless n was zero or a small negative number, since it would be pointing outside the valid bounds of the array's memory).

When we refer to *t[0] do we refer to int* p to the pointer itself because it is an element of the table or do we implicitly need a pointer to point at the slot than have the pointer point to the another place?

I'm not sure I understand this question. Since t[10] is an array of 10 pointers (ie 10 int * 's), that means that t[0] is a single item in that array and therefore has type int * . Therefore *t[0] dereferences the first pointer in the array, yielding the actual int value that the pointer is pointing to.

Answer 2

what int** means

It's the type of a pointer to a pointer to an int . If you dereference a variable t of this type (like this: *t ), you get a pointer to an int . If you dereference it twice (like this: **t ), you get an int .

Answer 3

If you have TYPE a[N]; , the array expression a , if evaluated, produces a pointer of type T * , pointing to a[0] . This is sometimes called C's array-to-pointer "decay" rule.

If TYPE is int * , as in your case, then TYPE * is int ** .

Since your array is of int * pointers, the pointer which indexes into the array is a necessarily pointer to that element type.