Differences when using ** in C

Question

I started learning C recently, and I\'m having a problem understanding pointer syntax, for example when I write the following line:

int ** arr = NULL;
         


        

Answer 1

The declaration int **arr says: "declare arr as a pointer to a pointer to an integer". It (if valid) points to a single pointer that points (if valid) to a single integer object. As it is possible to use pointer arithmetic with either level of indirection (i.e. *arr is the same as arr[0] and **arr is the same as arr[0][0]) , the object can be used for accessing any of the 3 from your question (that is, for second, access an array of pointers to integers, and for third, access an array of pointers to first elements of integer arrays), provided that the pointers point to the first elements of the arrays...

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Yet, arr is still declared as a pointer to a single pointer to a single integer object. It is also possible to declare a pointer to an array of defined dimensions. Here a is declared as a pointer to 10-element array of pointers to arrays of 10 integers:

cdecl> declare a as pointer to array 10 of pointer to array 10 of int;
int (*(*a)[10])[10]

In practice array pointers are most used for passing in multidimensional arrays of constant dimensions into functions, and for passing in variable-length arrays. The syntax to declare a variable as a pointer to an array is seldom seen, as whenever they're passed into a function, it is somewhat easier to use parameters of type "array of undefined size" instead, so instead of declaring

void func(int (*a)[10]);

one could use

void func(int a[][10])

to pass in a a multidimensional array of arrays of 10 integers. Alternatively, a typedef can be used to lessen the headache.