在C ++中使用'+'运算符将字符附加到字符串文字时会发生什么？

Question

Based on my reading, the following code:

string aggregate = "give" + 'n';

Should produce a resulting string with the value:

"given".

It instead produces garbage. Why doesn't the following happen?

1. "give" is converted to a std::string via the constructor that takes a pointer to a character array.

2. The '+' overload that takes a std::string and a character is called, returning a new string.

I am basing my theory on this man page.

Now, I have heard that the first argument to an overloaded operator isn't a candidate for constructor conversion if the operator is a member of a class. I believe I read that in Koenig and Moo. But, in this case I understand the '+' operator to be a non-member overload.

I realize this seems like a ridiculous over-complication, but I like to know FOR SURE what is happening when I write code.

Answer 1

The expression "give" + 'n' is evaluated first, adding (int)'n' to the address of the string constant "give" . The result of this pointer arithmatic is assigned to the string object. No error is raised because the result is of type const char* .

You should get used to thinking of everything on the right of the = as happening before the actual assignment (or initialization in this case).

You want the following:

// construct the string object first
std::string aggregate = "give";

// then append the character
aggregate += 'n';

or, explicitly build the a string object first:

std::string aggregate = std::string("give") + 'n';

Answer 2

The fact that there is a std::string on the left side of the expression doesn't matter for the first step: "give" is a const char[5] which decays to a const char* and 'n' is a char , that is, an integer type. Adding an integer to a pointer simply adds to the pointer's address. The result is again of type const char* , which is then implicitly converted to std::string .

Basically, your code is equivalent to:

const char* c = "a string" + 'n';
std::string s = c;

So, to achieve the desired effect, try

std::string s = std::string("a string") + 'n';

Answer 3

You're adding the integer 'n' to the address of the literal "give".

Try it with:

string aggregate = "long string long string long string long string long string long string long string " + 'A';

That should illustrate what's happening.

Answer 4

That code doesn't do what you think it does.

It takes the string literal "give" which then effectively degrades into a const char[5] , and then add the integer value of 'n' to that pointer. It then takes that new garbage pointer and tries to make a string out of it.

You want string aggregate = std::string("give") + 'n';