confusion in array of pointers

[nitishgupta]

char *name[]={"sachin","kap il","ajay"};
cout<<name[0]

what is the output of the above mentioned code?I think that
output must be an address because name[0]is storing the base address of sachin.But i get sachin as output.why is it so?
please help me!

[weaknessforcats]

You are correct. name[0] is an array. Therefore, he name of the array, name[0] in this case is the address of element 0 which is the "s" in "sachin. That is, name [0] is a char*.

However, The << operator of cout assumes that when a char* is to be displayed that it is a string and that you want to see the strng and not the address of the string.

To see the address of the string you will need to typecast name[0] to an int and then cout the int.