I stumbled upon "pointer to the entire array" and wrote a test program to clear some things out:
#include <stdio.h>
int main(){
int x[5] = {1, 2, 3, 4, 5};
int *a = x; // points to the 1st array element
int (*b)[5] = &x; // points to the 1st addres of the stream of 5 int elements
int (*c)[5] = x; // points to the 1st addres of the stream of 5 int elements
printf("%p\n%p\n%p\n\n",
(void *)a,
(void *)b,
(void *)c
);
++a;
++b;
++c;
printf("%p\n%p\n%p\n\n",
(void *)a,
(void *)b,
(void *)c
);
return 0;
}
This outputs:
0x7ffed0c20690
0x7ffed0c20690
0x7ffed0c20690
0x7ffed0c20694
0x7ffed0c206a4
0x7ffed0c206a4
To me it looks like lines:
int (*b)[5] = &x;
int (*c)[5] = x;
achieve exact same result, because:
they assign the same addres (in case of
*b
it is the address of entire array and in case of*c
it is the address of the first array member but those two overlap) andassign same pointer size of 5 ·
int
for*b
and*c
which leads to the exactly same pointer arithmetics when I increment the values.
Q1: Are there any hidden differences between definitions of *b
and *c
that I am missing?
Q2: Does pointer arithmetics only depends on the size of the pointer?
After you pointed I noticed that I do get an error:
main.c:9:16: warning: initialization of ‘int (*)[5]’ from incompatible pointer type ‘int *’ [-Wincompatible-pointer-types]
int (*c)[5] = x; // points to the 1st array element