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From: Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups: comp.lang.c
Subject: Re: One works, the other doesn't?
Date: Thu, 28 Nov 2019 15:54:15 -0800
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jak <nospam@please.ty> writes:

> Il 20/11/2019 12:14, Ben Bacarisse ha scritto:
>
>> jak <nospam@please.ty> writes:
>>
>>> Il 08/11/2019 01:02, Ben Bacarisse ha scritto:
>>>
>>>> jak <nospam@please.ty> writes:
>>
>> <cut>
>>
>>>>> typedef struct LIST
>>>>> {
>>>>>       char *str;
>>>>>       int count;
>>>>>       struct LIST *pMin, *pMaj;
>>>>> } Lista;
>>>>>
>>>>> Lista *BuildList(char *str)
>>>>> {
>>>>>       Lista *lRet = malloc(sizeof(Lista));
>>>>>       if(lRet != NULL)
>>>>>       {
>>>>>           lRet->str = malloc((strlen(str) + 1) * sizeof(str[0]));
>>>>
>>>> You could avoid doing two allocation by using a C99 flexible array
>>>> member:
>>>
>>> I know but unfortunately I can't afford to choose which compiler to
>>> use and during my work I meet really old compilers, so I tend to
>>> ignore the new features.
>>
>> There's a very old trick to do the same thing:
>>
>>    typedef struct LIST
>>    {
>>         int count;
>>         struct LIST *pMin, *pMaj;
>>         char str[1];
>>    } Lista;
>>
>> Then you over-allocate the struct and store the string in-place, so to
>> speak.  Of course it is technically undefined to access elements of the
>> str member with indexes > 1, but it was (and is) such a widely-used hack
>> that I would be surprised if any compiler did not take the possibility
>> into account.
>>
>> <cut>
>
> I guess I have to declare the string at the end of the struct.  [...]

You could use the original structure definition, and simply
point the 'str' member at the end of the struct (disclaimer:
not compiled):

    size_t extra = strlen( str ) + 1;
    Lista *p = malloc( sizeof *p  + extra );
    // check for allocation failure
    p->str = (char*)p + sizeof *p;

No hack required.