GLib atoms and memory chunks
-
24-09-2019 - |
문제
The following code snippet is from The Official GNOME 2 Developer's Guide:
GMemChunk my_chunk;
my_chunk = g_mem_chunk_new("My Chunk",
42,
42*16,
G_ALLOC_AND_FREE);
gchar *data[50000];
gint i;
/* allocate 40,000 atoms */
for(i = 0; i < 40000; i++)
{
data[i] = g_mem_chunk_alloc(my_chunk);
}
Does this mean every atom is of 42 bytes, each "memory chunk" contains
4216 atoms, and40000/16=2500
memory chunks will be created when the above code is run?Why are they using gchar* here? Does an implicit cast from gpointer (void*) to gchar* take place when
data[i] = g_mem_chunk_alloc(my_chunk);
is run?If the above statement is true then each gchar* points to 42 bytes of memory. How do I access all the bytes of a particular atom, then? Will
data[7]+41
be a usable memory location?
When I try to compile the code gcc produces this error message:
error: storage size of ‘my_chunk’ isn’t known
What's wrong?
해결책
In order of your questions:
- Each memory block contains 16 atoms, but other than that, yes.
- Yes,
void *
can be implicitly converted to any other pointer type in C, and this is typically considered to be good C style. They are usinggchar *
here because they apparently want to treat each atom as an array of 42gchar
s. - Yes,
data[7][41]
is the last accessible byte of the 8th atom. The error is because the declaration of
my_chunk
is wrong (GMemChunk
is an opaque type that shouldn't be directly instantiated in your code). The declaration should be:GMemChunk *my_chunk;
as per the signature of g_mem_chunk_new()
and g_mem_chunk_alloc()
.
By the way, the Glib documentation states that the Chunk allocator is deprecated, and you should use the slice allocator instead.