常量在Go中如何工作？

Question

In Go which snippet allocates less objects? Or do they both use the same amount of allocations, if so; why? (:

for i := 0; i < 10000000; i++ {
        log.println("hello")
    }

Does the code below allocates only 1 string?

const (
    HELLO string = "hello"
)

for i := 0; i < 10000000; i++ {
    log.println(HELLO)
}

Answer 1

Here's an adaptation of your program that prints out the underlying string representation in both cases.

package main

import (
    "fmt"
    "reflect"
    "unsafe"
)

const (
    Hello string = "hello"
)

func main() {
    for i := 0; i < 3; i++ {
        a := "hello"
        sh := (*reflect.StringHeader)(unsafe.Pointer(&a))
        fmt.Println(a, " ", *sh)
    }

    fmt.Println()

    for i := 0; i < 3; i++ {
        a := Hello
        sh := (*reflect.StringHeader)(unsafe.Pointer(&a))
        fmt.Println(a, " ", *sh)
    }
}

Here's the output:

hello   {4870353 5}
hello   {4870353 5}
hello   {4870353 5}

hello   {4870353 5}
hello   {4870353 5}
hello   {4870353 5}

The string header in {} in the output shows a pointer to the character data ("hello"), and the length of the string.

You can see that the pointer to the string data is the same throughout the program: the bytes data "hello" is referenced at exactly one memory address (here 4870353), no matter the loop count, and no matter whether it's a hardcoded string or a constant.

The language spec itself makes no guarantees about such behavior, but constant string interning is such a natural optimisation that it would be surprising if go behaved in any significantly different way.