What is the output of the following c++ code?

Question

char* a="coding";
char* b=(++a) + 3;
cout<<*b;

Answer: n

Can anyone explain how the output is n?

Answer 1

Yes I think you have encountered this question in the c++ challenges. As you know that the characters of "coding" are stored in contiguous memory locations. Initially "a" is a character pointer which points to the first memory location which is "c". As "a" gets incremented it points to "o" and when 3 is added to the pointer it moves three more steps further and points to "n".

Answer 2

a points to c as the first letter of coding. after incrementing it points to o and after plus 3 it points to n.

Answer 3

In c++ a string is not (only) an array. It is a standard class that has its own behavior. I am not a c++ expert but I guess that either cout or the <<-operator is overloaded for input of type string. Maybe some of the c++-experts here can clarify this.

Answer 4

Uppala Manoj When i tried cout<<a; then coding was printed and when i tried cout<<b; then ng was printed so it seems like when we cout without * then wherever the pointer is from there till the end character is printed.

But when i do this:
int g[5]={1,2,3,4,5};
int* h=g;
cout<<*h<<endl;
cout<<h<<endl;
then *h prints 1 so it's pointing to the first int here but h prints the address instead so why doesn't​ this happen with cout<<a; why does the whole coding gets printed instead of the address.