Strings

[vale minnow]

Hello,
when I initialize a new string, do I have to consider the last character '\0' by adding 1 to the length of the string ?
Imagine I want to initialize a string of 12 characters.
Do I have to do :
char mystring[12] or char mystring[13]
Thanks !

[frozen patrol]

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[ivory chasm]

char hello[5] = "hello"; // does not contain '\0'
char hello[6] = "hello"; // does contain '\0'
char hello[] = "hello"; // does contain '\0'
const char *hello = "hello"; // does contain '\0'

[lyric bough]

Whenever you use the string notation (quotation marks) then a bull t terminating byte is automatically appended.
If you use array notation then you need to manually append the newline.

So

char s[] = "ab";
``` is the same as:
```c
char s[] = { 'a', 'b', '\0' };

[late tree]

The first line compiles?!

char hello[5] = "hello";

It does T_T

[unkempt isle]

Compiler Output

<source>: In function 'main':
<source>:3:6: error: unused variable 'hello' [-Werror=unused-variable]
    3 | char hello[5] = "hello";
      |      ^~~~~
cc1: all warnings being treated as errors
Build failed

Kaznov#0698 | c | x86-64 gcc 13.1 | godbolt.org

[unkempt isle]

Compiler Output

<source>:3:6: error: unused variable 'hello' [-Werror,-Wunused-variable]
char hello[5] = "hello";
     ^
1 error generated.
Build failed

Kaznov#0698 | c | x86-64 clang (trunk) | godbolt.org

[late tree]

Neither GCC nor clang warn on it

They do however, if the array is even smaller - surprisingly it's still valid

[ivory chasm]

sometimes you want a non null terminated blob.

tbh this really isn't the way to do it... but

[vale minnow]

okkay tyvm !

[vale minnow]

okay thank u vm !

[frozen patrol]

@vale minnow Has your question been resolved? If so, run !solved :)

[vale minnow]

!solved