How to pass char** in python ctype?

C++, how to copy char pointer to char array using new producing different addresses?

• We have an exercise in our training that requires us to use the new operator to copy a char pointer to a char array. I have successfully done that by using a function. From main, a quoted string is passed to a function that accepts a char pointer. And I converted that char pointer to a string, and then converted it to a char array using new operator. I should catch that pointer in the main and then pass it again to the function. When I print the addresses of the char pointers, it is always the same. Should it be different because of the new operator? How can I make it different. Thank you. The functions goes like this: char* fCopyChar(char *oldchar) { string str( oldChar_cp ); char *newChar_cp = NULL; newChar_cp = new char[ str.size( ) + 1 ]; //copy( str.begin(), str.end(), newChar_cp ); int counter_i; for( counter_i =0; counter_i < str.size() ; counter_i++ ) { newChar_cp[ counter_i ] = *oldChar_cp; *oldChar_cp++; } newChar_cp[ str.size( ) ] = '\0'; cout << &newChar_cp << endl; cout << newChar_cp << endl; return newChar_cp; } int main( ) { char *copiedPointer_cp = NULL; copiedPointer_cp = fCopyChar( "This is a string." ); fCopyChar( copiedPointer_cp ); delete[] copiedPointer_cp; return 0; } Thank you again!

• Answer:

  The address of the pointers in the two calls relative to each other is indeterminent. It could be the same, or it could be different. It's likely to be the same because you are calling the same function(fCopyChar) twice from one function(main). So it's likely that the variables are located in the same spot on the stack frame both times. That's not the address of the array that's generated by 'new' though. The pointer holds an address, but it also has an address, which is the thing you are printing here:     cout << &newChar_cp << endl; I'm pretty sure that's not what you are trying to do. You are trying to get the address of the array created by 'new', right? To get the address of the array generated by 'new', normally you would just print the pointer, unadorned by any symbols like '&', like you're doing here:     cout << newChar_cp << endl; However, char pointers get special treatment when passed to printing functions, which print them as null terminated strings instead of addresses. To get the actual address then, you would want to cast it to a void pointer, like this:     cout << (void*)newChar_cp << endl;