Use of null character in strings (C++)

Question

I am brushing up on my C++ and stumbled across a curious behavior in regards to strings, character arrays, and the null character ('\0'). The following code:

#include <iostream>
using namespace std;

int main() {
    cout << "hello\0there"[6] << endl;

    char word [] = "hello\0there";
    cout << word[6] << endl;

    string word2 = "hello\0there";
    cout << word2[6] << endl;

    return 0;
}

produces the output:

> t
> t
>

What is going on behind the scenes? Why does the string literal and the declared char array store the 't' at index 6 (after the internal '\0'), but the declared string does not?

What happens when you `cout << word2`? That might give you insight into what is happening. — philipvr, Jul 20 '12 at 15:30
@philipvr printing all 3 of the representations produces simply the string "hello". — ewok, Jul 20 '12 at 15:33
If he were to `cout << word2` it would print out `hello`, which does not really shed light onto his problem. — sean, Jul 20 '12 at 15:33

score 9 · Accepted Answer · edited Jul 20 '12 at 15:38

9

From what I remember, the first two are in essence just an array and the way a string is printed is to continue to print until a \0 is encounterd. Thus in the first two examples you start at the point offset of the 6th character in the string, but in your case you are printing out the 6th character which is t.

What happens with the string class is that it makes a copy of the string into it's own internal buffer and does so by copying the string from the start of the array up to the first \0 it finds. Thus the t is not stored because it comes after the first \0.

edited Jul 20 '12 at 15:38

ewok

20,148
51
149
254

answered Jul 20 '12 at 15:30

sean

3,955
21
28

2

Also, in the case of `string`, you are indexing out of the boundaries of the `string`'s memory. You are lucky your code doesn't crash. – Shahbaz Jul 20 '12 at 15:32
12

As a side note, if he really does want the whole string copied he could use `std::string word3("hello\0there", 11);` – GWW Jul 20 '12 at 15:32
@Shahbaz unlucky you mean, it doesn't help to understand to let undefined behavior :[ – kingsjester Sep 29 '22 at 07:44

score 6 · Answer 2 · answered Jul 20 '12 at 15:32

Because the std::string constructor that takes a const char* treats its argument as a C-style string. It simply copies from it until it hits a null-terminator, then stops copying.

So your last example is actually invoking undefined behaviour; word2[6] goes past the end of the string.

score 4 · Answer 3 · answered Jul 20 '12 at 15:30

4

You are constructing a string from a char* (or something that decayed to that). This means that the convention for C-strings apply. That is they are '\0' terminated. That's why word2 only contains "hello".

answered Jul 20 '12 at 15:30

RedX

14,749
1
53
76

score 1 · Answer 4 · edited May 23 '17 at 12:17

The problem is that you are not printing strings at all - you are printing single characters.

char word [] = "hello\0there";//Array of char...
cout << word[6] << endl;      //So word[6] is the char't' (NOT a string)

string word2 = "hello\0there"; //std::string...
cout << word2[6] << endl;      //so word2[6] is the char 't' (NOT a string as well)

So, you are invoking the "char" overloads, not the "char*" or "string" overloads at all, and the NULL chars have nothing to do with it at all : You are just printing the 6th character of word, and the 6th character of word2.

If I am reading your intent correctly, your test should read:

cout << &(word[6]) (char*, should print "there")
cout << &(word2[6]) (char* as well, undefined behaviour pre-C++11)

In C++11 and later this will also print "there" And be well defined

Use of null character in strings (C++)

4 Answers4

Linked