We known that the * operator is used for declaration of pointer and we can only put the address in the pointer eg int *p=&a;

Question

why this code running
char*c="c+"; //pointer only takes address
cout<<c<<endl;
c++;
cout<<c;
/*output
c+
+
*/

Answer 1

pointer can be used to declare arrays and strings and it always points to the first element in that array or string.

Answer 2

Yes, we pass to cout, but if cout sees it is a pointer to a character array (a.k.a string) it will print that string. If you want to determine how a pointer should be shown you can use printf(), you can print its address with something like %d, the first character with %c and the whole string with %s.

Answer 3

Because it is a pointer to a two character array. When you increment the pointer you move it to the next value in the array.

Answer 4

thanks #vlad and #mazen...

Answer 5

if its pointer to char array then we will only pass the address to the pointer but in this program we r passing the string that contains two character..