Accessing Characters by Using Bracket Notation

With a standard C-style string, you can use brackets to access individual characters:

char city[40] = "Amsterdam";
cout << city[0] << endl; // display the letter A

In C++ the two bracket symbols constitute a single operator, the bracket operator, and you can overload this operator by using a method called operator[](). Typically, a binary C++ operator (one with two operands) puts the operator between the two operands, as in 2 + 5. But the bracket operator places one operand in front of the first bracket and the other operand between the two brackets. Thus, in the expression city[0], city is the first operand, [] is the operator, and 0 is the second operand.

Suppose that opera is a String object:

String opera("The Magic Flute");

If you use the expression opera[4], C++ looks for a method with this name and signature:

String::operator[](int i)

If it finds a matching prototype, the compiler replaces the expression opera[4] with this function call:

opera.operator[](4)

The opera object invokes the method, and the array subscript 4 becomes the function argument.

Here’s a simple implementation:

char & String::operator[](int i)
{
    return str[i];
}

With this definition, the statement

cout << opera[4];

becomes this:

cout << opera.operator[4];

The return value is opera.str[4], or the character 'M'. So the public method gives access to private data.

Declaring the return type as type char & allows you to assign values to a particular element. For example, you can use the following:

String means("might");
means[0] = 'r';

The second statement is converted to an overloaded operator function call:

means.operator[][0] = 'r';

This assigns 'r' to the method’s return value. But the function returns a reference to means.str[0], making the code equivalent to

means.str[0] = 'r';

This last line of code violates private access, but because operator[]() is a class method, it is allowed to alter the array contents. The net effect of the code is that "might" becomes "right".

Suppose you have a constant object:

const String answer("futile");

Then, if the only available definition for operator[]()is the one you’ve just seen, the following code is labeled an error:

cout << answer[1];  // compile-time error

The reason is that answer is const, and the method doesn’t promise not to alter data. (In fact, sometimes the method’s job is to alter data, so it can’t promise not to.)

However, C++ distinguishes between const and non-const function signatures when overloading, so you can provide a second version of operator[]() that is used just by const String objects:

// for use with const String objects
const char & String::operator[](int i) const
{
    return str[i];
}

With the definitions, you have read/write access to regular String objects and read-only access to const String data:

String text("Once upon a time");
const String answer("futile");
cout << text[1];    // ok, uses non-const version of operator[]()
cout << answer[1];  // ok, uses const version of operator[]()
cin >> text[1];     // ok, uses non-const version of operator[]()
cin >> answer[1];   // compile-time error