Golang anonymous struct初始化指针切片

Question

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?

Answer 1

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.

Answer 2

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