access elements of string after `\0`

Question

Recently I have seen a code in which a string was being sorted, and \0 (since it has smallest ASCII code) came first of the string.

But using format specifier %c the true sorted string was printed, but I checked with %s and nothing was printed.

So here is my question, if I add a \0 in a string (while I have valid value in my string after terminator), could I access those values with something like %c, is it safe? And will this answer always or it just answered because of UB?

Here is an example:

int main()
{
    char a[8]="abcdefg";
    char b[8];
    for (int i = 0; i < 8; i++)
    {
        b[7-i] = a[i];
    }
    for (int i = 0; i < 8; i++)
        printf("%c ", b[i]);
    printf("%s", b);
}

In above example printf("%c ", b[i]); prints the hole string , even after \0.

In second one printf("%s", b); nothing is printed?

Actually I want to know if using first printf is true and legal?

Does this answer your question? [How do I print a non-null-terminated string using printf?](https://stackoverflow.com/questions/2137779/how-do-i-print-a-non-null-terminated-string-using-printf) — GSerg, Apr 04 '20 at 12:33
@GSerg no,sorry for unclear question, I add an example of to clarify the question. — hanie, Apr 04 '20 at 12:47
@hanie It pretty much does. You are using `%c` to print a single character in a `for` loop, as opposed to having it output the entire string at once which would not be right. — GSerg, Apr 04 '20 at 12:52
@GSerg yet I don't understand is using this `printf("%c ", b[i]);` right work to do.I mean it don't have any risk? — hanie, Apr 04 '20 at 12:56

Ardent Coder · Accepted Answer · 2020-04-04T13:10:16.453

char a[8]="abcdefg";

Note: a[7] contains '\0' (automatically added).

char b[8];
for (int i = 0; i < 8; i++)
{
    b[7-i] = a[i];
}

Note: b is the reverse of a => b[0] contains the null character ('\0')

Now, getting into your questions:

for (int i = 0; i < 8; i++)
    printf("%c ", b[i]);

This prints b character by character, but not as an entire string. There is no problem in doing this.

And the output will look something like this:

<empty> g f e d c b a

That empty space is because b[0] is a null character.

in second one printf("%s", b); nothing is printed?

Again, b[0] contains the null character i.e. the first character of this string is \0. So it is treated as an empty string though it contains a text. The null character marks the end of a string.

actually I want to know if using first printf is true and legal?

Of course, that can be seen as printing a char array. Not every char array has to be a "string"!

score 1 · Answer 2 · edited Apr 04 '20 at 13:32

This solution uses a +1 to skip the first unwanted '\0' terminator and the precision part of the format specifier to make sure to only print the desired string length (which I assume to be known, since the size of the array is fixed).

#include <stdio.h>

int main()
{
    char a[8]="abcdefg";
    char b[8];
    for (int i = 0; i < 8; i++)
    {
        b[7-i] = a[i];
    }
    printf("%.7s", b+1);
}

This works because in the call to printf() the parameter b is treated like a pointer to char and a +1 skips the first of those chars, which happens to be the undesired terminator.

Then the '%s' is used, with extended options. The precision specifier .7 set up now minimum width (i.e. no padding) by the empty left side of the . and a maximum of 7 characters to be printed, by the part right of the ..

This way it is possible to print the string with %s and drop the terminator.

In case the length is dynamic (e.g. you want to use the initially unknown length of a) you can upgrade to the next level of complexity and use the *version of the precision. That would allow to put the "7" into a variable instead.

:-) Saw the typo myself, but thought "Naaah. Somebody is bount to be already editing..." — Yunnosch, Apr 04 '20 at 13:33
Haha, I just gained the editing privilege so thought of showing it off lol — Ardent Coder, Apr 04 '20 at 13:35

score 0 · Answer 3 · answered Apr 04 '20 at 12:32

0

A string is just an array of characters. So as long as you don't go out of bounds of the array, everything is well defined.

This of course requires you to know the "capacity" of the string.

answered Apr 04 '20 at 12:32

Šimon Tóth

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"A string is just an array of characters" is more like "A string is just an array of characters with a null character at the end" as C defines a _string_ as "A string is a contiguous sequence of characters terminated by and including the first null character." – chux - Reinstate Monica Apr 04 '20 at 13:47

access elements of string after `\0`

3 Answers3