Why it is possible to do
const string exclam = "!"; const string str = exclam + "Hello" + " world"; And not possible to do this:
const string exclam = "!"; const string str = "Hello" + " world" + exclam; I know (although can't understand why) that it is not allowed to do:
const string str = "Hello" + " world" + "!"; as it will be interpreted like const char[6] + const char[6] + const char[1], so from other side, why this is not allowed also, or why it uses char[] and not string.
The + operator is left-associative (evaluated left-to-right), so the leftmost + is evaluated first.
exclam is a std::string object that overloads operator+ so that both of the following perform concatenation:
exclam + "Hello" "Hello" + exclam Both of these return a std::string object containing the concatenated string.
However, if the first two thing being "added" are string literals, as in:
"Hello" + "World" there is no class type object involved (there is no std::string here). The string literals are converted to pointers and there is no built-in operator+ for pointers.
It's because you are concatanating const char[6] + const char[6], which is not allowed, as you said.
In C++, string literals (stuff between quotes) are interpreted as const char[]s.
You can concatenate a string with a const char[] (and vice-versa) because the + operator is overridden in string, but it can't be overridden for a basic type.
const char[N] where N is the size of the string literal. It is an array, and like an array it can be implicitly converted to a pointer to its first element (a const char*). In C++, while the conversion of a string literal to (non-const) char* is deprecated, it is still allowed.const string exclam = "!"; // Initialize a c++ string with an ansi c string const string str = exclam + "Hello" + " world"; // Using the operator+ method of exclam You can do it because the operator+ of exclam will return a new string containing "!Hello", on which you subsequently call the operator+ with " world" as parameter, so you get another string which, finally, gets assigned to str by means of the copy constructor.
On the other hand
const string str = "Hello" + " world" + exclam; cannot be executed because "Hello" is just a const char *, which doesn't have a operator+ taking a const char * as parameter.
(New answer, this was not possible back in 2010)
You can now write
const string str = "Hello"s + " world"s + "!"s; // ^ ^ ^ By adding that s after a string literal, you tell the compiler it's actually a std::string and not a const char[]. That means you can call members functions on it. E.g. ("ABC"s).back() but also +
In addition to the answers that explain the reason for your observations, I post here how to get rid of the problem (you might have figured this out already).
Replace
const string str = "Hello" + " world" + exclam; with
const string str = string("Hello") + " world" + exclam; so you make sure the first operand is a string and not a const char[].
+:const string str = "Hello" " world" + exclam;