toStringz or not toStringz

[Regan Heath]

On Tue, 12 Jul 2011 17:09:04 +0100, Steven Schveighoffer  
<schveiguy at yahoo.com> wrote:

> On Tue, 12 Jul 2011 11:41:56 -0400, Regan Heath <regan at netmail.co.nz>  
> wrote:
>
>> On Tue, 12 Jul 2011 15:59:58 +0100, Steven Schveighoffer  
>> <schveiguy at yahoo.com> wrote:
>>
>>> On Tue, 12 Jul 2011 10:50:07 -0400, Regan Heath <regan at netmail.co.nz>  
>>> wrote:
>>
>>>>> What if you expect the function is expecting to write to the buffer,  
>>>>> and the compiler just made a copy of it?  Won't that be pretty  
>>>>> surprising?
>>>>
>>>> Assuming a C function in this form:
>>>>
>>>>    void write_to_buffer(char *buffer, int length);
>>>
>>> No, assuming C function in this form:
>>>
>>> void ucase(char* str);
>>>
>>> Essentially, a C function which takes a writable  
>>> already-null-terminated string, and writes to it.
>>
>> Ok, that's an even better example for my case.
>>
>> It would be used/called like...
>>
>>    char[] foo;
>>    .. code which populates foo with something ..
>>    ucase(foo);
>>
>> and in D today this would corrupt memory.  Unless the programmer  
>> remembered to write:
>
> No, it wouldn't compile.  char[] does not cast implicitly to char *.   
> (if it does, that needs to change).

Replace foo with foo.ptr, it makes no difference to the point I was making.

>> I am assuming also that if this idea were implemented it would handle  
>> things intelligently, like for example if when toStringz is called the  
>> underlying array is out of room and needs to be reallocated, the  
>> compiler would update the slice/reference 'foo' in the same way as it  
>> already does for an append which triggers a reallocation.
>
> OK, but what if it's like this:
>
> char[] foo = new char[100];
> auto bar = foo;
>
> ucase(foo);
>
> In most cases, bar is also written to, but in some cases only foo is  
> written to.
>
> Granted, we're getting further out on the hypothetical limb here :)  But  
> my point is, making it require explicit calling of toStringz instead of  
> implicit makes the code less confusing, because you understand "oh,  
> toStringz may reallocate, so I can't expect bar to also get updated" vs.  
> simply calling a function with a buffer.

This is not a 'new' problem introduced the idea, it's a general problem  
for D/arrays/slices and the same happens with an append, right?  In which  
case it's not a reason against the idea.

>>>> You might initially extern it as:
>>>>
>>>>    extern "C" void write_to_buffer(char *buffer, int length);
>>>>
>>>> And, you could call it one of 2 ways (legitimately):
>>>>
>>>>    char[] foo = new char[100];
>>>>    write_to_buffer(foo, foo.length);
>>>>
>>>> or:
>>>>
>>>>    char[100] foo;
>>>>    write_to_buffer(foo, foo.length);
>>>>
>>>> and in both cases, toStringz would do nothing as foo is zero  
>>>> terminated already (in both cases), or am I wrong about that?
>>>
>>> In neither case are they required to be null terminated.
>>
>> True, but I was outlining the worst case scenario for my suggestion,  
>> not describing the real C function requirements.
>
> No, I mean you were wrong, D does not guarantee either of those (stack  
> allocated or heap allocated) is null terminated.  So toStringz must add  
> a '\0' at the end (which is mildly expensive for heap data, and very  
> expensive for stack data).

Ah, ok, this was because I had forgotten char is initialised to 0xFF.  If  
it was initialised to \0 then both arrays would have been full of null  
terminators.  The default value of char is the killing blow to the idea.

>>> The only thing that guarantees null termination is a string literal.
>>
>> string literals /and/ calling toStringz.
>>
>>> Even "abc".dup is not going to be guaranteed to be null terminated.   
>>> For an actual example, try "012345678901234".dup.  This should have a  
>>> 0x0f right after the last character.
>>
>> Why 0x0f?  Does the allocator initialise array memory to it's offset  
>> from the start of the block or something?
>
> The final byte of the block is used as the hidden array length (in this  
> case 15).

Good to know.

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