Initializing a char * with an expression does not work

Question

The following code is producing the incorrect output:

string my_string="My_First_Text";
char * my_pointer=(char *)(my_string+"My_Second_Text").c_str();

Why? As I am initializing my_pointer, I presumed that no my_pointer=new char[100] is needed. If this assumption is not true, then why?

Answer 1

Note that my_string+"My_Second_Text" is a temporary std::string, which will be destroyed after the expression immediately. That means my_pointer will become dangled at once, because the char array it's supposed to point to has been destroyed along with the destroy of the temporary std::string; note that the returned char array belong to std::string, it's not standalone. Then, deference on dangled pointer would lead to UB.

string my_string="My_First_Text";
char * my_pointer=(char *)(my_string+"My_Second_Text").c_str();
// the temporary constructed from my_string+"My_Second_Text" has been destroyed
// my_pointer is dangled now; deference on it is UB

Using named variable instead of temporary will be fine. e.g.

string my_string = "My_First_Text";
my_string += "My_Second_Text";
const char * my_pointer = my_string.c_str();

BTW: The return type of std::basic_string::c_str is const char*, and any modification on it is UB. So trying to convert it to char* explicitly is dangerous.

Writing to the character array accessed through c_str() is undefined behavior.