About pointers in C?

hi..uhmm..about pointers in C.





char *s;


s="hello";





when u print the *s, the output is garbage or nonesense..


but if u print the value of s alone...hello would appear..why is this so? correct me if im wrong...


the s pointer points to the address of "hello" and when i dereference 8 it should work..by now...

About pointers in C?
I think you're doing it wrong.





You would need something like this:





#include %26lt;iostream%26gt;


using namespace std;





void main()


{


char *s;


char mystring[] = "hello";





s = mystring;





cout %26lt;%26lt; s;


system("Pause");


}





EDIT - I just tried this and it works, have a good day.
Reply:it won't print garbage .


it will just print h;


the pointer s will be just pointing to the first element .


so when you do a *s it will print h
Reply:its like this:





s stored the address location


*s stores the value stored in the first position of s


for printf and all functions that use string (i.e. char *) values, you pass the address and all manipulation etc happens to the value stored at address using address pointer -%26gt; pointer manipulation





when u print as:


printf("%s", *s); --%26gt; this should print junk





but when u print as:


printf("%c", *s); --%26gt; this should print the char at first byte location of s





and if u print as:


printf("%s", s); --%26gt; this should print the entire string stored in s





read a book on pointers, it will take too long to explain here. and its already explained very well in many books.





also there is a chance your current program (as it is) may core dump. this is because u r defining in one statement and then assigning in another. but since you did not allocate memory before proceeding to assign, your program will core dump. to resolve this, use:





char *s="hello";





doing this does definition, allocation of memory and assignation all at the same time.