英文:
Golang anonymous struct in initializing slice of pointers
问题
从《GOPL》第7章(第7.6节)中,我注意到了这一行代码:
var tracks = []*Track{
{"Go", "Delilah", "From the Roots Up", 2012, length("3m38s")},
{"Go", "Moby", "Moby", 1992, length("3m37s")},
{"Go Ahead", "Alicia Keys", "As I Am", 2007, length("4m36s")},
{"Ready 2 Go", "Martin Solveig", "Smash", 2011, length("4m24s")},
}
我对如何初始化 Track
指针的切片感到困惑。
所以后来我尝试了以下示例:
type Ex struct {
A, B int
}
a := []Ex{Ex{1, 2}, Ex{3, 4}}
b := []Ex{{1, 2}, {3, 4}}
c := []*Ex{&Ex{1, 2}, &Ex{3, 4}}
d := []*Ex{{1, 2}, {3, 4}}
e := []*Ex{{1, 2}, &Ex{3, 4}}
我发现这5种情况都没有错误。{1, 2}
似乎是一个名为 "匿名结构体" 的结构体,但它让我感到困惑,因为即使我尝试填充 *Ex
指针而不是 Ex
结构体,它也能正常工作。
此外,当我尝试以下代码时,它会报语法错误:
f := []*Ex{&{1, 2}, &{3, 4}} // 语法错误!
有人可以帮忙解释一下这些情况实际上是怎么回事吗?
英文:
From Chapter 7 of GOPL (Section 7.6), I noticed this line:
var tracks = []*Track{
{"Go", "Delilah", "From the Roots Up", 2012, length("3m38s")},
{"Go", "Moby", "Moby", 1992, length("3m37s")},
{"Go Ahead", "Alicia Keys", "As I Am", 2007, length("4m36s")},
{"Ready 2 Go", "Martin Solveig", "Smash", 2011, length("4m24s")},
}
I was kind of confused by how it initialized the slice of Track
pointers.
So later I tried the following example:
type Ex struct {
A, B int
}
a := []Ex{Ex{1, 2}, Ex{3, 4}}
b := []Ex{{1, 2}, {3, 4}}
c := []*Ex{&Ex{1, 2}, &Ex{3, 4}}
d := []*Ex{{1, 2}, {3, 4}}
e := []*Ex{{1, 2}, &Ex{3, 4}}
I found all of the 5 cases are okay with no errors. The {1, 2}
seems to be a structure called "anonymous struct", but it confuses me since it works fine even when I am trying to fill in *Ex
pointers instead of Ex
struct.
Furthermore, when I try the following code, it complains syntax error:
f := []*Ex{&{1, 2}, &{3, 4}} // Syntax Error!
Can somebody help explain what is actually going on in these cases?
答案1
得分: 6
我认为"anonymous"在这里并不是完全正确的词。这个结构体有一个名字,"Track"或者"Ex"。但是你用一种简化的方式进行了初始化:
f := []<type>{{...}, {...}}
大致上等同于:
f := []<type>{<type>{...}, <type>{...}}
但是它比简单的字符串替换要聪明一些。如果<type>
是一个指针类型,比如*Ex
或者*Track
,它也会自动正确地进行初始化:
f := []*<type>{{...}, {...}} // 等同于...
f := []*<type>{&<type>{...}, &<type>{...}}
对于映射类型也是一样的:
f := map[string]*<type>{"a": {...}, "b": {...}} // 等同于...
f := map[string]*<type>{"a": &<type>{...}, "b": &<type>{...}}
进一步澄清一下,匿名结构体是指没有单独的类型定义的结构体。这些是匿名类型,但不是匿名结构体。匿名结构体是一个没有关联类型定义的结构体。初始化值时不需要引用类型的语法对于使其易于管理是至关重要的:
f := []struct{
A, B int
}{
{1, 2}, {3, 4}
}
// 或者使用指针...
f := []*struct{
A, B int
}{
{1, 2}, {3, 4}
}
在列表中的结构体没有类型定义,因此它是匿名的。由于初始化的简化语法,这是可以的 - 我不需要一个名字来初始化它。
英文:
I think anonymous isn't quite the correct word here. The struct has a name, "Track" or "Ex". But you are initializing it with a shortcut:
f := []<type>{{...}, {...}}
is roughly the same as:
f := []<type>{<type>{...}, <type>{...}}
But it's a little smarter than blind string replacement. If <type>
is a pointer like *Ex
or *Track
, it also automatically initializes correctly:
f := []*<type>{{...}, {...}} // is the same as...
f := []*<type>{&<type>{...}, &<type>{...}}
It works the same for maps:
f := map[string]*<type>{"a": {...}, "b": {...}} // is the same as...
f := map[string]*<type>{"a": &<type>{...}, "b": &<type>{...}}
For further clarification, anonymous structs are ones that have no separate type definition. These are anonymous types, but not anonymous structs. An anonymous struct is a struct with no associated type definition. This syntax of initializing values without referring to the type is essential to making the manageable:
f := []struct{
A, B int
}{
{1, 2}, {3, 4}
}
// or with pointers...
f := []*struct{
A, B int
}{
{1, 2}, {3, 4}
}
Here the struct within the list has no type definition, so it is anonymous. And thanks to the initialization shorthand, that's fine - I don't need a name to initialize it.
答案2
得分: 0
请详细说明一下你对Ex结构的定义。
但是,如果Ex被实现为以下形式:
type Ex struct {
t interface{}
o interface{}
}
你没有给出匿名结构体的指针,因为定义是未知的。
此外,你不能这样做:
variable := new({1, 2})
// 或者
variable := {1, 2}
当你这样做时:
b := []Ex{{1, 2}, {3, 4}}
// 你创建了一个包含两个Ex结构的切片
// 所以 b := []Ex{Ex{t:1,o:2}, Ex{t:3, o:4}}
相反地:
f := []*Ex{&{1, 2}, &{3, 4}}
// 你创建了一个未知定义类型的切片
// f = []Ex{Ex{t: addressOfUnknowType, o: addressOfUnknowType}}
希望对你有所帮助。
英文:
Please give more detail on your structure definition of Ex.
But, if Ex is implemented as this:
type Ex struct {
t interface{}
o interface{}
}
you don't give a pointer of anonymous struct because, the definition it's not know.
Also you can't do:
variable := new({1, 2})
// or
variable := {1, 2}
When you do:
b := []Ex{{1, 2}, {3, 4}}
// you create slice of Ex with two Ex struct
// so b := []Ex{Ex{t:1,o:2}, Ex{t:3, o:4}}
contrariwise:
f := []*Ex{&{1, 2}, &{3, 4}}
// you create slice of unknow definition type
// f = []Ex{Ex{t: addressOfUnknowType, o: addressOfUnknowType}}
I hope help you
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