Go version
go1.22
Output of go env in your module/workspace:
What did you do?
Compile the following program:
package main
type key struct {
u16 uint16
u8 bool
u32 uint32
}
var map1 map[key]struct{}
var map2 map[uint64]struct{}
var key1 = key{}
var key2 = uint64(0)
var sink struct{}
func main() {
sink = map1[key1]
sink = map2[key2]
}What did you see happen?
The compiler output:
0x0023 00035 (main.go:17) CALL runtime.mapaccess1(SB)
...
0x0040 00064 (main.go:18) CALL runtime.mapaccess1_fast64(SB)
What did you expect to see?
runtime.mapaccess1_fast64 being used on both since the key type is smaller than a uint64.
That said, there is some complications. For example, through unsafe, someone could writes values to the padding in-between the bool field and the uint32. Also, someone using unsafe could write a bit pattern to the bool that isn't 0 or 1.
It isn't clear to me that this is something that someone can rely on.
However, even if we make the key struct exactly 64-bits wide with no padding and no bools, it still doesn't use runtime.mapaccess1_fast64.
Go version
go1.22
Output of
go envin your module/workspace:What did you do?
Compile the following program:
What did you see happen?
The compiler output:
What did you expect to see?
runtime.mapaccess1_fast64being used on both since thekeytype is smaller than a uint64.That said, there is some complications. For example, through
unsafe, someone could writes values to the padding in-between theboolfield and theuint32. Also, someone usingunsafecould write a bit pattern to the bool that isn't 0 or 1.It isn't clear to me that this is something that someone can rely on.
However, even if we make the
keystruct exactly 64-bits wide with no padding and no bools, it still doesn't useruntime.mapaccess1_fast64.