Array and pointers to BDSM

Look at the code first:

1#include <stdio.h>2 3 intMainvoid)4 {5     intI= About;6     int*p = &i;7p[0] = the;8printf"What's in it?%d\n", p[0]);9printf"What's ' s ' i ' now?%d\n", i);Ten     return 0; One}

Look at the results again:
What's in it? 88
What's ' s ' I ' now? 88
That is, when you point the pointer p to the memory address of I, the behavior of P can behave like an array.

Assigning a value to P[0] is a number 88 for that address and three contiguous memory units behind it.

Because the int type occupies 4 bytes, and one memory unit corresponds to one byte.

The value of I is also the same address, that is, the address of p[0], so the value of I is rewritten.

If you assign a value of p[1], the value of I does not change. But actually assigning a value to P[1] Returns a segment error.

* It is estimated that the memory address is not allowed to write. But the compiler will not find this error.

Array and pointers to BDSM