What is the difference between these two cases of adding a string?

Question

I noticed that when I initialized a string, the compiler reported an error that I was not expecting.

For example:

#include <iostream>
#include <string>

using namespace std;

int main() {
    string s1 = "Hello", s2 = "World!"; // ok
    string s3 = s1 + ", " + "World!"; // ok
    string s4 = "Hello" + ", " + s2; // error
    cout << s1 + " " + s2 << endl; //ok
    return 0;
}

For me, if s3 worked just fine, s4 should do the same.

Why do I get that error? What is the difference between these two initialization strings (s3 and s4)?

Answer 1

"Hello" is not a std::string (rather it is a const char[6], while ", " is a const char[3]) and the + operator for std::strings does not apply.

This is a minor inconvenience of C++ and stems from its C ancestry. It means that in concatenating strings via + you must ensure that at least one of its two operands actually is a std::string, as in

auto s = std::string{"Hello"} + ", " + "World";

where the first + has a std::string as its first operand and hence produces a std::string, so that the second + also has a std::string as its first operand (since + is processed from left to right).

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Edit1 prompted by the comment by T.C., I mention that C-string literals are automatically concatenated if only separated by white space:

std::string s = "Hello" ", " "World";

This behaviour too is inherited from C: the preprocessor renders above code to

std::string s = "Hello, World";

before the compiler proper processes it (to be more precise, string concatenation takes place in phase 6 of translation, just before compilation). This is in fact the most simple and hence most convenient way to concatenate raw string literals. But note that I had to declare the type of s, since auto deduction would have given a const char*.

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Edit2 prompted by the comment by PaperBirdMaster, I mention that since C++14, you can directly form std::string literals by simply adding s after the string, if you pull in the associated operator""s (or surrounding namespace).

using std::literals::operator""s;                // pull in required operator
const auto s = "Hello"s + ", " + "World";        // std::string

See this post as to why the required operator""s is hidden in the nested namespace. Note also that alternatively to using std::literals::operator""s; you may pull in the surrounding namespace: either of the following declarations will do.

using namespace std::string_literals;
using namespace std::literals::string_literals;
using namespace std::literals;

which are not as bad (and damned) as plain using namespace std;.

Answer 2

The following line attempts to add 2 pointers together.

string s4 = "Hello" + ", " + s2; // error

"Hello" is a const char[6] which decays to const char* and ", " is a const char[3] which decays to const char*.

You are trying to assign the addition of two pointers to a std::string, which is not valid. Furthermore, the addition of two pointers is not valid either.

As other people have said, this doesn't have anything to do with the overloads of operator+ of the std::string class, as the following code would also not compile for the same reason:

string s4 = "Hello" + ", ";

I have added some clarifications provided by user @dyp.

You can fix your problem by using the C++ Standard Library "user-defined" string literal. The functionality is provided by std::literals::string_literals::operator""s which is found in the <string> header. You might have to use a using-directive or -declaration; this was not required on my compiler and including the <string> header was enough. Example:

string s4 = "Hello"s + ", "s + s2;

This is the equivalent of:

string s4 = std::string("Hello") + std::string(", ") + s2;

When you place an instance of std::string as the first operand, the rest of the expression will use that type, because of left to right associativity for operator+ will make sure that an instance of std::string is returned as the left-most operand every time, instead of const char*. That is why you have no error on line:

string s3 = s1 + ", " + "World!";

That line is equivalent to:

string s3 = ((s1 + ", ") + "World!");