pointers

[ibtkm]

hi every body.
i have one fundamental question
in this code :
const char* a[2] = {"ab","bc"};
"a" array is a array of pointers but why we get it "ab" and "bc" instead of address?

[R0mai]

Because in the expression :
cout << a[0] operator<<() is overloaded explicitly for char*.

[ibtkm]

if it's overloaded for char* so why when we write :

char a[3] = "ab";
char b[3] = "cd";
const char* names[4] = {a,b};
cout << names[0]; 

answer is "ab" instead of address of a[0];

[rahulroot]

well your are confusing yourself with array of pointers.
names is array of pointers to char.

so names[0] type is const char *,once you gives this type of datatype to any ostream or printf,it will print the data pointed by that pointer,in this case it will print ab.