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Character Array, \0





I know, the \0 on the end of the character array is a must if you use the character array with functions who expect \0, like cout, otherwise unexpected random characters appear.

My question is, if i use the character array only in my functions, reading it char by char, do i need to store the \0 at the end?

Also, is it a good idea to fill only characters and leave holes on the array?

Consider the following:

char chars[5];

chars[1] = 15;
chars[2] = 17;
chars[3] = 'c';

//code using the chars[1] and chars[3], but never using the chars
int y = chars[1]+chars[3];
cout << chars[3] << " is " << y;

Does the code above risk unexpected errors?

EDIT: edited the example.

like image 260
SkyRipper Avatar asked Mar 21 '23 09:03


2 Answers

The convention of storing a trailing char(0) at the end of an array of chars has a name, it's called a 'C string'. It has nothing to do, specifically, with char - if you are using wide character, a wide C string would be terminated with a wchar_t(0).

So it's absolutely fine to use char arrays without trailing zeroes if what you are using is just an array of chars and not a C string.

char dirs[4] = { 'n', 's', 'e', 'w' };
for (size_t i = 0; i < 4; ++i) {
    fprintf(stderr, "dir %d = %c\n", i, dirs[i]);
    std::cout << "dir " << i << " = " << dirs[i] << '\n';

Note that '\0' is char(0), that is it has a numeric, integer value of 0.

char x[] = { 'a', 'b', 'c', '\0' };

produces the same array as

char x[] = { 'a', 'b', 'c', 0 };

Your second question is unclear, though

//code using the chars[1] and chars[3], but never using the chars
int y = chars[1]+chars[3];
cout << chars[3] << " is " << y;

Leaving gaps is fine, as long as you're sure your code is aware that they are uninitialized. If it is not, then consider the following:

char chars[4]; // I don't initialize this.
chars[1] = '1';
chars[3] = '5';
int y = chars[1] + chars[3];
std::cout << "y = " << y << '\n';
// prints 100, because y is an int and '1' is 49 and '5' is 51

// later
for (size_t i = 0; i < sizeof(chars); ++i) {
    std::cout << "chars[" << i << "] = " << chars[i] << '\n';


char one = 1;
char asciiCharOne = '1';

are not the same. one has an integer value of 1, while asciiCharOne has an integer value of 49.

Lastly: If you are really looking to store integer numeric values rather than their character representations, you may want to look at the C++11 fixed-width integer types in . For an 8-bit, unsigned value uint8_t, for an 8-bit signed value, int8_t

like image 65
kfsone Avatar answered Apr 01 '23 12:04


Running off the end of a character array because it has no terminating \0 means accessing memory that does not belong to the array. That produces undefined behavior. Often that looks like random characters, but that's a rather benign symptom; some are worse.

As for not including it because you don't need it, sure. There's nothing magic that says that an array of char has to have a terminating \0.

like image 44
Pete Becker Avatar answered Apr 01 '23 10:04

Pete Becker