Question

In the following code:

int main()
{
   char names[2][11] = {"Manchester","Party"};
   char (*jk)[11];
   jk = names;                    // LINE 1
   char gaming[10] = {"Jetking"};
   char (*po)[10];
   po = &gaming;                  // LINE 2
   cout<<"PO is "<<*po;

Line 2 requires me to put & in front of gaming, while Line 1 doesnt. The error it gives for Line 2 when I dont put & is, “error: cannot convert ‘char [10]’ to ‘char (*)[10]’ in assignment” ? I didnt quite understand this part. Since “char (*po)[10];” can be interpreted as a pointer to an array of 10 characters.

Answer 1

In the assignment

jk = names;

the array names of type char[2][11] is converted to a pointer to its first element, thus it decays into a char (*)[11].

gaming, however is converted into a char* in most contexts, and that char* is incompatible with the type that po has, char (*)[10], so that assignment is invalid. If you take the address of gaming, you get exactly the required char (*)[10].