whats the answer and how does it work?

Question

Char *p="sololearn" ;
cout<<p+1;

Answer 1

̶Y̶o̶u̶r̶ ̶c̶h̶a̶r̶a̶c̶t̶e̶r̶ ̶t̶y̶p̶e̶ ̶p̶o̶i̶n̶t̶e̶r̶ ̶i̶n̶i̶t̶i̶a̶l̶l̶y̶ ̶p̶o̶i̶n̶t̶s̶ ̶t̶o̶ ̶t̶h̶e̶ ̶f̶i̶r̶s̶t̶ ̶c̶h̶a̶r̶a̶c̶t̶e̶r̶ ̶i̶n̶ ̶t̶h̶e̶ ̶s̶t̶r̶i̶n̶g̶.̶ ̶T̶h̶e̶n̶ ̶y̶o̶u̶ ̶i̶n̶c̶r̶e̶m̶e̶n̶t̶ ̶i̶t̶ ̶b̶y̶ ̶o̶n̶e̶,̶ ̶s̶o̶ ̶t̶h̶e̶ ̶o̶u̶t̶p̶u̶t̶ ̶w̶i̶l̶l̶ ̶b̶e̶ ̶'̶o̶'̶.̶

Answer 2

Shouldn't it be cout<<*(p+1);?

Answer 3

If you type cout<<p it will print the whole word, and if you add 1, like p+1 the word will start from the second letter (o) in this case

Answer 4

Write it like this *(p+n--)
And the explanation for your case is that it takes the ASCII value of 's' because thats *p and adds 2 on it, so the output is 117

Answer 5

No problem

Answer 6

@Filip halp me on this one =^=

Answer 7

Ohhh, with the explanation?

Answer 8

Yes, *p prints just one character, and p prints the whole word, because you wrote it like
char *p="sololearn";

Answer 9

I've explained it all in code, so if you want check it out. It's related to your questions:
https://code.sololearn.com/cz9JBW1SOgvB/?ref=app

Answer 10

but when it's cout<<*(p+1) it's just o ??

Answer 11

thank you

Answer 12

but the output is ololearn according to compiler

Answer 13

no just cout <<p+1

Answer 14

yes please I don't how it works

Answer 15

last question I initialize a variable INT n=2;
cout<<*p+n--;
output is 117 ??

Answer 16

wow I understood completely