Thread: terminating null

today we were talking about arrays
my teacher said that C only counts \0 as 1 byte - not 2 (as i thought),
is this true, or does he need to read up on C more often ?

He is indeed correct the character '\0' is called the null termination character, (ASCII code 0) the \ means the next character in line (in this case 0) is an escape character type, not the actual number zero.

Null termination is used in character arrays to determine if the array is to be treated as a string or not.

Code:

char str[] = {'H','e','l','l','o' };

This character array contains no null termination character, so C sees this as a simple array of characters, each its own entity(if you will...).

Code:

char str[] = {'H','e','l','l','o','\0' };

This character array contains the same letters plus the extra null termination character, this to C is the string literal "Hello".

EDIT: Also note, that when loading a character array with a string you don't need to load it as i have done above

Code:

char str[] = "Hello";

This will work jsut fine and C will add the null termination character for you.

There are several escape characters just like this

Code:

\n = newline
\t = tab
\" = Quotes in a String Literal
\' = Single Quote
\\ = '\' character
\b = backspace
etc...

there are several more, any standard C/C++ book will describe them to you.

An important note on the null termination character -

You will notice that the array without the \0 in it is only 5 characters long, where as the array with the \0 is 6. This is important to note becuase if you are making a string you must make sure that it is big enough to house all of the information in the literal + 1 space for the null termination character, or you will get a buffer overflow error.

Code:

char str[5] = "Hello";

ERR this will cause an overflow error.

Code:

char str[6] = "Hello";

There we go, this will work, but remember you don't have that extra space to fill with anything, becuase the null termination character already took up that spot.


Listen to your teachers, they are right *most* of the time

You will notice that the array without the \0 in it is only 5 characters long, where as the array with the \0 is 6. This is important to note becuase if you are making a string you must make sure that it is big enough to house all of the information in the literal + 1 space for the null termination character, or you will get a buffer overflow error.

Code:

char str[5] = "Hello";

ERR this will cause an overflow error.

That's not an overflow error, the null character is just discarded. In fact, this is a useful trick if you want an array of characters but not a string and you're not keen on using the verbose array initialization. If you want a string, it's easier to let the array size itself:

Code:

char str[] = "Hello";

Listen to your teachers, they are right *most* of the time

Yea, way to make him doubt his teacher. How about "Listen to your teachers unless you have a good reason to doubt them".

Originally Posted by bithub

He didn't say it was an overflow error, he said it would cause an overflow error which is certainly true the next time any string operation is performed on it. If you name a char array str, then it sure as hell better be null terminated.

The results is implementation dependent.. VC++ 6.0 produces this error and will not allow you to do that.

Code:

C:\dvlp\test1\test1.cpp(9) : error C2117: 'Hello' : array bounds overflow

Dev-C++ does not produce and compile-time error or warning.

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int i;
    char str[5] = "Hello";
    int n = strlen(str);
    printf("len = %d\n",n);
    for(i = 0; i < n; i++)
        printf("%d ", str[i]);
    printf("%d\n");
    system("pause");
    return 0;
}

Here is the output from the above, and the output is different
each time I run the program. It cleary illustrates that str[] is not null terminated, as Narf said in an earlier post.

Code:

len = 9
72 101 108 108 111 96 -3 127 25 25
Press any key to continue . . .

He didn't say it was an overflow error, he said it would cause an overflow error which is certainly true the next time any string operation is performed on it.

Ooh, semantics. By that argument, you can say that the other example will also cause an overflow error. Which is certainly true the next time you use an out of bounds subscript. Now, you can be willfully stupid and treat the array as if it's a string when you know it's not, and yes it will break, or you can use it as an array.

You obviously haven't been here long enough.

Just because one teacher doesn't know C from a hole in the ground doesn't mean that all teachers are as dumb. If you don't trust your teacher, get another one. If you don't trust any teachers, don't you think you're a bit too cynical?

On a broader note, you should not automatically believe what your instructors say just because they are teaching it. Be critical, probe for reasons and answers. If nothing else it'll help you understand the material better.

There's a difference between healthy amount of skepticism and thinking all teachers are stupid. You can either try to learn the language as it's taught and correct bad habits later, or spend all of your time trying to prove your teacher wrong and learn nothing because you're trying to swallow everything all at once. That's my opinion, and I'm sticking to it.

Originally Posted by Charmy

Code:

char str[] = {'H','e','l','l','o','\0' };

This character array contains the same letters plus the extra null termination character, this to C is the string literal "Hello".

That's not a string literal. It's simply an array of characters. You cannot modify string literals. Even this isn't a string literal:

Code:

char foo[] = "not a string literal";

This is a string literal:

Code:

char *foo = "is a string literal";

You can edit the individual characters in the first example, not in the second.

Originally Posted by Narf

Just because one teacher doesn't know C from a hole in the ground doesn't mean that all teachers are as dumb. If you don't trust your teacher, get another one. If you don't trust any teachers, don't you think you're a bit too cynical?

Go read a Schildt book. It's not cynicism, it's healthy paranoia.


Quzah.

Originally Posted by Ancient Dragon

The results is implementation dependent.. VC++ 6.0 produces this error and will not allow you to do that.

Code:

C:\dvlp\test1\test1.cpp(9) : error C2117: 'Hello' : array bounds overflow

Since you are posting on the C board, why don't you compile that as C, not as C++? You might get the same error, but I don't have VC, so I can't confirm.

Originally Posted by swoopy

Since you are posting on the C board, why don't you compile that as C, not as C++? You might get the same error, but I don't have VC, so I can't confirm.

You're right -- I use *.c file and it compiled without errors or warnings. It produced similar test results that I got with Dev-C++ .

>It produced similar test results that I got with Dev-C++ .
Which is what we expected. This is a case where it would be nice to get some warning from the compiler. Perhaps Dev-C++ may warn with the -Wall or -pedantic switch.