pointers - 在 Go 中取消引用指针

Question

我对为什么第 15 行无效感到困惑。为什么不能取消引用指向 big.Int 的指针，而指向 int 的指针可以？

package main

import (
    "fmt"
    "big"
)

func main() {
    var c *int = getPtr()
    fmt.Println(c)
    fmt.Println(*c)

    var d *big.Int = big.NewInt(int64(0))
    fmt.Println(d)

    // does not compile - implicit assignment of big.Int
    // field 'neg' in function argument
    //fmt.Println(*d)
}

func getPtr() *int {
    var a int = 0
    var b *int = &a
    return b
}

Answer 1

这是因为 Int 是具有未导出字段的结构。当您按值将结构传递给函数时，您正在制作它的副本。 Go spec声明这是合法的

...either all fields of T must be exported, or the assignment must be in the same package in which T is declared. In other words, a struct value can be assigned to a struct variable only if every field of the struct may be legally assigned individually by the program.