Why the output is 'c', can anyone say me?

Question

const char *p="12345";
const char **q=&p;
*q="abcde";
const char *s=++p;
p="UVXYZ";
cout<<*++s;

Answer 1

0. p holds address of constant "12345"
1. q holds address of p.

So now *q = p = &("12345")

2. Change *q to hold address of constant "abcde"

This also replaces p, so now p holds the same address [&("abcde")].

3. Increment p to address of constant "bcde"
4. Assign s = p = address of constant "bcde"
5. Change p to hold address of constant "UVXYZ"
6. Increment s to address of constant "cde"
7. Output the const char pointed to by s, which is  'c'.

Answer 2

You are welcome! 🤝

Answer 3

Ok thank you Brian

Answer 4

Nikhil Maroju, Step 5 begins with both p and s holding duplicate copies of the memory address where "bcde" is located. Then p gets reassigned to hold a different address where "UVXYZ" is stored. The address stored in s does not change, and the constant that it points to does not change.

Answer 5

Hey thank you Brian ,
But at 5th step we are changing p to hold address of constant "UVWYZ" so 's' value should be changed right!, As s is holding same address