>> That there is an indirection in the struct doesn't mean that a function
doesn't need to know the size of the memory it is allocating for a bucket.
Your judgement are now incorrect.
Obtaining the size of the of the type is not so hard at runtime.
It always known from `rtype`
This is how map allocates buckets.
It creates an array of the type `t.bucket`.
Where `t` is `mapType` and the `bucket` is a field in this type which
specifies the the better type of storing buckets.
if h.buckets == nil {
h.buckets = newarray(t.bucket, 1)
}
How new array know the size?
This is easiest question in the world.
Answer is: From a given type.
// implementation of make builtin for slicesfunc newarray(typ *_type, n
uintptr) unsafe.Pointer {
flags := uint32(0)
if typ.kind&kindNoPointers != 0 {
flags |= flagNoScan
}
if int(n) < 0 || (typ.size > 0 && n > _MaxMem/uintptr(typ.size)) {
panic("runtime: allocation size out of range")
}
return mallocgc(uintptr(typ.size)*n, typ, flags)
}
Look how passed the `key` into the `map` method `get` (mapaccess).
It passed as the unsafe.Pointer.
func mapaccess1(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
P.S.
The `map` is very well oprimized in Go language (it has more than one method
for different groups of types).
But this does not means that the code generated for use "a single method for
every groups of types" approach would be much more slower.
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