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From: Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups: comp.lang.c
Subject: Re: C String terminator
Date: Tue, 21 Jan 2020 04:10:18 -0800
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Xref: csiph.com comp.lang.c:149303

Lew Pitcher <lew.pitcher@digitalfreehold.ca> writes:

> Two minor corrections
>
> Lew Pitcher wrote:
>
>> T wrote:
>>
>>> On 2020-01-17 12:48, T wrote:
>>>
>>>> On 2020-01-17 12:46, Ben Bacarisse wrote:
>>>>
>>>>> T <T@invalid.invalid> writes:
>>>>>
>>>>>> On 2020-01-16 15:45, Ben Bacarisse wrote:
>>>>>>
>>>>>>> T <T@invalid.invalid> writes:
>>>>>>>
>>>>>>>> On 2020-01-16 12:17, Ben Bacarisse wrote:
>>>>>
>>>>> <cut>
>>>>>
>>>>>>>>> $ gcc -o t t.c
>>>>>>>>> $ ./t
>>>>>
>>>>> <cut>
>>>>>
>>>>>>>> I am a total idiot when it comes to C.  How do
>>>>>>>> I run this?
>>>>>>>
>>>>>>> I don't know.  That depends on your compiler, OS and so on and that
>>>>>>> obviously different to mine sice whole process of compiling and
>>>>>>> running is shown above.
>>>>>>
>>>>>> I have Fedora Linux at my disposal, which has
>>>>>> a gcc c compiler.
>>>>>
>>>>> Then I am baffled.  I showed both the gcc command to compile the code,
>>>>> and the command to run it.  I've removed everything else so you can see
>>>>> them.
>>>>>
>>>>> <cut>
>>>>
>>>> Ben!  I am an idiot when it comes to C.  You
>>>> got to start out like I am a total noob!
>>>
>>> Okay, I looked at it from a bash standpoint instead
>>> of a mystery language I had no clue about.  And I
>>> feel dumb.  You were very precise!
>>>
>>> $ cd C
>>> $ vi t.c
>>> $ gcc -o t t.c
>>> $ t
>>> [FOOp@]
>>>
>>>
>>> Question:  why did I get
>>>      [FOOp@]
>>>
>>> and you got
>>>      [FOO]
>>> ?
>>
>> Because, as implied by Ben's source code comment, the program given
>>
>>>>>>>>> #include <stdio.h>
>>>>>>>>>
>>>>>>>>> void foo(const char *s)
>>>>>>>>> {
>>>>>>>>>     printf("[%s]\n", s);
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> int main(void)
>>>>>>>>> {
>>>>>>>>>      char str[3] = "FOO";
>>>>>>>>>      foo(str); // DANGER, WILL ROBINSON!
>>>>>>>>> }
>>
>> invokes undefined behaviour.
>>
>> It does this by passing an array of three characters (ultimately) to
>> printf(), which /expects/ a C "string" as it's first argument.  The array
>
> Correction #1: in this case, the critical point is that, here, printf()
> expects a C "string" as it's second argument (as well as the first argument,
> which is the given format string "[%s]\n").

Technically, that isn't right.  A %s conversion specifier requires
as its argument a pointer to the first element of an array of
character type, not a pointer to a string.  It is true in this
particular case that the array must have a null character in it to
avoid undefined behavior, but the argument description says "pointer
to array of character type", and does not mention "string".

(Also:  the possessive of "it" is "its", with no apostrophe.  The
word "it's", with an apostrophe, is a contraction of "it is".)