What is the difference between '\0' and "\0" in C++ string

Question

I tried to add a null terminator (\0) in a string in two ways. One using '\0' and another using "\0", and I'm getting different outputs in both cases:

#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    string test1 = "";
    test1 += "\0";
    cout << test1.size() << endl;
    
    string test2 = "";
    test2 += '\0';
    cout << test2.size() << endl;
}

I am getting output

0
1

Can anyone please explain why the string is taking '\0' as a character not a null terminator? If I print test1 or test2, it will print an empty string.

Answer 1

'\0' is a single character. A char with value 0.

"\0" is an array of two characters. A const char[2] with value { 0, 0 } (The first is the one specified in the code for the string literal, and the second is the nul terminator added by the compiler to all string literals).

When you have test2 += '\0';, this passes the one character to the string object to append to it. The function called is std::string::operator+=(char).

When you have test2 += "\0";, this passes a pointer to the array's 1st element to the string object, calling std::string::operator+=(const char*). This is interpreted as a C-string, which can't have embedded nul characters. There is no way to tell between "" and "\0" when you only have a pointer, since the first character is '\0' in both cases and that looks like the end of the string.

If you want to use embedded nul characters, you have to pass the size separately. This can be done with std::string::append(), eg:

// ptr+size
test2.append("\0", 1);

// iterator pair
const auto& str = "\0";
test2.append(std::begin(str), std::end(str)-1);  // (-1 for the extra nul terminator)

Or, using a string_view, which carries its size with it:

using namespace std::literals::string_view_literals;
test2 += "\0"sv;
// "\0"sv.size() == 1