Get end of string as const char * in C++

Question

Is it correct that in order to get a char pointer to the end of a string in C++ i have to do this:

std::string str = ...
const char * end_ptr = &*str.cend();

Like dereferencing, then getting the address of the expression. Is there a more elegant way to do this?

Answer 1

I don't think that this is guaranteed, I'd use

const char* end_ptr = str.c_str() + str.size();

to get a pointer to the terminating '\0'.

Answer 2

You solution is not correct. It will lead to undefined behavior. You cannot dereference str.cend():

Returns an iterator to the character following the last character of the string. This character acts as a placeholder, attempting to access it results in undefined behavior.

You can do

const char* end_ptr = &str[str.size() - 1];

As pointed out in comment, this will be one bit short.

You can use std::c_str to match the exact behavior you want and get a pointer to the \0:

Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.

The pointer is such that the range [c_str(); c_str() + size()]

Then you just have to

const char* end_ptr = str.c_str() + str.size();
// or in c++11 you can use data()

Assuming you have at least one character in your string.