英文:
Assign value to parent golang structure field
问题
有两种情况：
第一种情况：
type A struct {A_FIELD string}type B struct {AB_FIELD string}func main() {b := &B{A_FIELD: "aaaa_field",B_FIELD: "bbbb_field",}}
第二种情况：
type A struct {A_FIELD string}type B struct {AB_FIELD string}func main() {b := &B{}b.A_FIELD = "aaaa_field"b.B_FIELD = "bbbb_field"fmt.Printf("Good!")}
为什么第二种情况可以正常工作，而第一种情况不行？我收到了编译时异常。我应该如何修改第一种情况使其正常工作？
英文:
There are two situations:
type A struct {A_FIELD string}type B struct {AB_FIELD string}func main() {b := &B{A_FIELD: "aaaa_field",B_FIELD: "bbbb_field",}}
And
type A struct {A_FIELD string}type B struct {AB_FIELD string}func main() {b := &B{}b.A_FIELD = "aaaa_field"b.B_FIELD = "bbbb_field"fmt.Printf("Good!")}
Why the second one is working, but the first one is not? I receive compile time exceptions. How should I change first one to work?
答案1
得分: 7
为什么第二个能够工作，而第一个不能工作？
因为
b.A_FIELD = "aaaa_field"
实际上是
b.A.A_FIELD = "aaaa_field"
的伪装。
我应该如何修改第一个才能使其工作？
func main() {b := &B{A: A{A_FIELD: "aaaa_field",},B_FIELD: "bbbb_field",}}
你应该阅读《Effective Go》中的嵌入的工作原理。
英文:
> Why the second one is working, but the first one is not?
Because
b.A_FIELD = "aaaa_field"
is actually
b.A.A_FIELD = "aaaa_field"
in disguise.
> How should I change first one to work?
func main() {b := &B{A: A{A_FIELD: "aaaa_field",},B_FIELD: "bbbb_field",}}
You should read how embedding work on Effective Go.
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