go 1.11.1 source code is below:
Generally speaking, make chan just pay attention to the presence or absence
of buf.
When I saw the source code of make chan, I can understand case 1: chan buf
is 0, but can't understand case 2 & default.
Who knows this principle?
Thanks!
var c *hchan
switch {
case size == 0 || elem.size == 0:
// Queue or element size is zero.
c = (*hchan)(mallocgc(hchanSize, nil, true))
// Race detector uses this location for synchronization.
c.buf = c.raceaddr()
case elem.kind&kindNoPointers != 0:
// Elements do not contain pointers.
// Allocate hchan and buf in one call.
c = (*hchan)(mallocgc(hchanSize+uintptr(size)*elem.size, nil, true))
c.buf = add(unsafe.Pointer(c), hchanSize)
default:
// Elements contain pointers.
c = new(hchan)
c.buf = mallocgc(uintptr(size)*elem.size, elem, true)
}
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