std::string getting (char *) instead of (const cha

Question

std::string.c_str() returns a (const char *) value. I Googled and found that I can do the following:

std::string myString = "Hello World";
char *buf = &myString[0];

How is this possible? &myString[0] is an object of type std::string, so how can this work?

Answer 1

There are const and non-const overloads of std::string::operator[]( size_type pos ), and the non-const version returns a char&, so you can do things like

std::string s("Hello");
s[0] = 'Y';

Note that, since s[0] returns char&, then &s[0] is the address of element s[0]. You can assign that address to a char*. It is up to you not to misuse this pointer.

Answer 2

It has to do with operator precedence. The [] operator has higher precedence than the address-of operator &, so the address-of operator works on the character reference returned by the strings operator[] function.

Answer 3

The std::string methods c_str() and operator[] are two diferent methods, which return two different types.

The method c_str() does indeed return a const char*.

const char* c_str() const;

However, the method operator[] returns instead a reference to a char. When you take the address of it, you get the address of a char.

       char& operator[] (size_t pos);
 const char& operator[] (size_t pos) const;

Answer 4

The std::string type has an operator[] that allows indexing each one of the characters in the string. The expression myString[0] is a (modifiable) reference to the first character of the string. If you take the address of that you will get a pointer to the first character in the array.

Answer 5

You are wrong. &myString[0] is not of type std::string, it is of type char *.

The [] operator has higher precedence and operator[] returns a reference to char, and its address (the & operator) is of type char *.

Answer 6

The operator [] (std::string::char& operator[] (size_t pos)) overloaded returns a reference to the character at the index. You are taking the address of such character reference which is fine.

So, myString[0] return type is not std::string but char&.

There is no reason to do it. You can directly do -

myString[0] = 'h';