将'\\ 0'添加到std :: string时出现意外行为

Question

Why does the C++ standard allow the following?

#include <iostream>
#include <string>    

int main()
{
    std::string s(10, '\0'); // s.length() now is 10
    std::cout << "string is " << s << ", length is " << s.length() << std::endl;
    s.append(5, '\0'); // s.length() now is 15 
    std::cout << "string is " << s << ", length is " << s.length() << std::endl;
    // the same with += char and push_back 

    // but:
    s += "hello"; // s.length() returns 20 string is "hello"
    std::cout << "string is " << s << ", length is " << s.length() << std::endl;

    return 0;
}

Why does it add 0 and count it? It looks like broken integrity of string, doesn't it? But I checked standard and it is correct behavior.

Answer 1

Why does standard allows following?

Because the people designing C++ strings decided that such things should be allowed. I'm not sure if anyone that was part of the team that designed C++ strings are on SO... But since you yourself say that the standard allows it, that's the way it is, and I doubt it's about to change.

It's sometimes quite practical to have a string that can contain "anything". I can think of a few instances when I've had to work around the fact that C style strings can't contain zero-bytes. Along with the fact that long C style strings take a long time to find the length of, the main benefit of C++ strings is that they are not restricted to "what you can put in them" - that's a good thing in my book.

Answer 2

Not sure what is problem here.

Adding '\\0' in the middle of the std::string changes nothing - null character is treated like any other. The only thing that can change is if you use .c_str() with function that accepts null-terminated strings. But then it's not problem of .c_str() , only with the function that treats '\\0' specially.

If you want to know how many characters has this string as if treated like null-terminated string, use

size_t len = strlen(s.c_str());

Note that it's O(n) operation, because that's how strlen works.

If you ask why += operator doesn't add the implicit null character of string literal "hello" to the string, I say the reverse (adding it) is unclear and definitely not what you want 99% of the time. On the other hand, if you want to add '\\0' to your string, just append it like a buffer:

char buffer[] = "Hello";
s.append(buffer, sizeof(buffer));

or (even better) drop the char arrays and null-terminated strings altogether and use C++-style replacements like std::string as NTS-replacement, std::vector<char> as contiguous buffer, std::vector as dynamic array with pointers replacement, and std::array (C++11) as standard C array replacement.

Also, (as mentioned by @AdamRosenfield in comments), your string after adding "hello" does have in fact 20 characters, it's probably only that your terminal doesn't print nulls.

Answer 3

NUL char '\\0' is the ending character for c style string , not std::string s. However, it supports this character to get values from a const char pointer so that it can find the end of a c-style string. Otherwise, it is treated just like other characters

Answer 4

std::string is more of a container for characters than anything else and \\0 is a character. As a real world example, take a look at the CreateProcess function in Windows. The lpEnvironment parameter takes a null-terminated block of null-terminated strings (ie A=1\\0B=2\\0C=3\\0\\0 ). If you're building a block it's convenient to use an std::string.