If you just want to pass a std::string
to a function that needs const char*
you can use
std::string str;
const char * c = str.c_str();
If you want to get a writable copy, like char *
, you can do that with this:
std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0
// don't forget to free the string after finished using it
delete[] writable;
Edit: Notice that the above is not exception safe. If anything between the new
call and the delete
call throws, you will leak memory, as nothing will call delete
for you automatically. There are two immediate ways to solve this.
boost::scoped_array
boost::scoped_array
will delete the memory for you upon going out of scope:
std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0
// get the char* using writable.get()
// memory is automatically freed if the smart pointer goes
// out of scope
std::vector
This is the standard way (does not require any external library). You use std::vector
, which completely manages the memory for you.
std::string str;
std::vector<char> writable(str.begin(), str.end());
writable.push_back('\0');
// get the char* using &writable[0] or &*writable.begin()
Read it backwards (as driven by Clockwise/Spiral Rule):
int*
- pointer to int
int const *
- pointer to const int
int * const
- const pointer to int
int const * const
- const pointer to const int
Now the first const
can be on either side of the type so:
const int *
== int const *
const int * const
== int const * const
If you want to go really crazy you can do things like this:
int **
- pointer to pointer to int
int ** const
- a const pointer to a pointer to an int
int * const *
- a pointer to a const pointer to an int
int const **
- a pointer to a pointer to a const int
int * const * const
- a const pointer to a const pointer to an int
- ...
And to make sure we are clear on the meaning of const
:
int a = 5, b = 10, c = 15;
const int* foo; // pointer to constant int.
foo = &a; // assignment to where foo points to.
/* dummy statement*/
*foo = 6; // the value of a can“t get changed through the pointer.
foo = &b; // the pointer foo can be changed.
int *const bar = &c; // constant pointer to int
// note, you actually need to set the pointer
// here because you can't change it later ;)
*bar = 16; // the value of c can be changed through the pointer.
/* dummy statement*/
bar = &a; // not possible because bar is a constant pointer.
foo
is a variable pointer to a constant integer. This lets you change what you point to but not the value that you point to. Most often this is seen with C-style strings where you have a pointer to a const char
. You may change which string you point to but you can't change the content of these strings. This is important when the string itself is in the data segment of a program and shouldn't be changed.
bar
is a constant or fixed pointer to a value that can be changed. This is like a reference without the extra syntactic sugar. Because of this fact, usually you would use a reference where you would use a T* const
pointer unless you need to allow NULL
pointers.
Best Solution
That means you are trying to call non-const member function
addDoctor
from a const member function.EDIT: In the newly posted code, you have an array of
Doctor
objects but you are trying to put an pointer into that array. That will give an error like can not convert fromDoctor*
toDoctor