I NEED EXPLAIN!

Question

const char *p="12345";
const char **q=&p;
*q="abcde";
const char *s=++p;
p="XYZWVU";
cout<<*++s;
THE OUTPUT:c.

Answer 1

const char *p="12345"; // same as const char p[]="12345"
const char **q=&p; // pointer to p[]
*q="abcde"; // p[] = "abcde"
const char *s=++p; // p = &p[1]; s == &p[1]
p="XYZWVU"; // trash address at p
cout<<*++s; // s == &p[2]; output p[2] or 'c'

Answer 2

Because the data is the constant not the pointer to it. This protects both:
const char * const p = "12345";

Answer 3

John Wells: Why did const not protect the char* string literal?

Answer 4

p holds the address of your string. You told it hold "XYXWVU" instead killing the address plus two bytes beyond.

Answer 5

John Wells ,

Thank you for the information.

Answer 6

John Wells why p is trash address ?