C++ string and overloaded operators

Question

I'm studying about the C++ library type string. I was looking into operation that the type can perform, among which are concatenation. I know that in C++ operators can be overloaded to suit class types needs,and I know that character string literals are const char arrays under the hood. So, I know that the string type defines a constructor that takes a string literal argument and converts it into a string, so that I can initialize a string using:

string s="hello!"

what happens here is an impicit conversion (through the constructor that takes a const char*) to string and then a copy initialization is performed (through the copy-constructor of string).

so far so good.. (correct me if I am getting something wrong so far)

Now, I know that I can concatenate two strings, or I can concatenate a string and a char literal (or vice versa) and a string and a string literal (so I suppose the first happens because there's an overloaded + operator that takes symmetrically a char and a string, the latter happens thanks to the overloaded + opeartor that concatenates two strings and the implicit conversion from const char* to string). My questions are:

1: why can't I concatenate two string literals? shouldn't the operator+ that takes two strings be called (and two implicit conversion performed from const char* to string) ?

string s="hi"+" there";

2: I tried to concatenate a char literal and a string literal (calling the operator+ that takes a char and a string) but I get only garbage in the resulting string, as the compiler doesn't mind, but the result is certainly not what I want.

string s='h'+"ello!";
cout<<s<<endl;

I get garbage out.If a operator+(char,const string&) exists it should be called after implicitly converting "ello" to string shouldn't it?

can you please explain?

Answer 1

1) Concatenating two string literals.

"Hello" + " world"

The compiler has no hint that it should be looking for anything related to std::string -- it is looking at two const char[] objects, and you cannot + those (or the const char * they could be degraded to).

2) Using operator+ on a char literal and a string literal.

(Thanks to User dyp for pointing in the right direction.)

If you literally mean a char literal -- eg 'a' -- then you're running afoul of one of the more surprising things C/C++ have to offer.

Consider: a[i] is equivalent to i[a] . (This is a fact.)

So, if you write:

'a' + "Hello"

...that is equivalent to (in ASCII)...

"Hello" + 97

...which is a pointer into nowhere, ie constructing a std::string from it is undefined behaviour.

Answer 2

Borgleader already gave half the answer in comment:

1. when adding two string litterals, compiler only sees two const char * and they could be cast to string but also to integers. So unless you explicitely say that you want them as strings, you get an error

2. when adding a string litteral and a char, compiler sees a pointer and an integral number . It knows how to process that and does pointer arithmetics on your C string - which is probably not what you expected! There are chances that you end past the end of the char array and invoke undefined behaviour. Examples:

    "abc" + '0' => "abc" + 48 : 44 positions past the NULL ... "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghij" + '0' => "ij"