On 6/8/2011 12:13 PM, Josh Gargus wrote:
On Jun 8, 2011, at 11:56 AM, Kevin Jones wrote:
Comments and explanations inline below.
--- On *Wed, 6/8/11, Julian Leviston /<[email protected]
<mailto:[email protected]>>/* wrote:
Also, what is this?
struct vtable *_vt[0];
Is that creating a pointer to a struct array which has zero
elements in it? I don't follow what that's for?
This is an idiom in C for variable-length arrays when at the end
of a struct definition.
Thanks Kevin,
Can you elaborate a bit more on this idiom? Is it purely
intention-revealing code, to signify that it's a pointer to an array
of structs instead of a single struct? In other words, does it
compile to the same thing as "struct vtable *_vt;" ?
nope...
it only states where the start of the array is, in the struct, but no
actual storage is reserved in the struct (any array contents are assumed
to exist following the end of the struct).
typedef struct Foo_s Foo;
struct Foo_s {
int x;
float y[0];
};
Foo *p;
p=malloc(sizeof(Foo)+10*sizeof(float)); //allocate struct, + the size
of 10 floats.
p->y[5]; //accesses the float at element 5, which is past the end of
the struct
p->y[-1]; //happens to be at the same address as p->x on typical systems
...
to really understand this though would require to understand pointers,
and how C lays things out in memory.
say, p==0x006F4000
and addresses:
0x006F4000 p->x
0x006F4004 p->y[0]
0x006F4008 p->y[1]
0x006F400C p->y[2]
...
or, laid out as bytes:
<XX XX XX XX> <Y0 Y0 Y0 Y0> <Y1 Y1 Y1 Y1> <Y2 Y2 Y2 Y2>
or, if p->x=0x12345678, p->y[0]=1.0; p->y[1]=0; and using little-endian
ordering:
<78 56 34 12> <00 00 80 3F> <00 00 00 00> ...
dunno if this helps...
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