英文:
How is it that I can initialize a slice of pointers like this?
问题
我很惊讶我可以用这种方式初始化指针的切片：
package main
import (
"fmt"
)
type index struct {
i, j int
}
func main() {
indices := []*index{{0, 1}, {1, 3}} // 为什么这样可以工作？
fmt.Println(*indices[1])
}
我原本以为我需要写得更冗长一些，像是：
indices := []*index{&index{0, 1}, &index{1, 3}}
我在文档中可以找到这个内容吗？
英文:
I was surprised I could initialize a slice of pointers in this way:
package main
import (
"fmt"
)
type index struct {
i, j int
}
func main() {
indices := []*index{{0, 1}, {1, 3}} // Why does this work?
fmt.Println(*indices[1])
}
I was expecting to have to write something more verbose like:
indices := []*index{&index{0, 1}, &index{1, 3}}
Where would I find this in the documentation?
答案1
得分: 5
从规范中可以看到：
> 在类型为T的数组、切片或映射的复合字面值中，如果元素或映射键本身是复合字面值，并且其类型与T的元素或键类型相同，则可以省略相应的字面值类型。类似地，如果元素或键是复合字面值的地址，并且元素或键类型是*T，则可以省略&T。
基本上，它已经知道每个元素将是*index，所以它省去了你重复输入&index的麻烦。
如果切片类型与元素类型不同（例如它是接口类型的切片），你需要像这样指定每个元素的类型：
indices := []interface{}{&index{0, 1}, &index{1, 3}}
英文:
From the spec:
> Within a composite literal of array, slice, or map type T, elements or
> map keys that are themselves composite literals may elide the
> respective literal type if it is identical to the element or key type
> of T. Similarly, elements or keys that are addresses of composite
> literals may elide the &T when the element or key type is *T.
Basically, it already knows each element will be a *index, so it saves you having to actually type &index over and over again.
If the slice type is not the same as the element type (perhaps it is a slice of an interface type), you would have to specify the type of each element like so:
indices := []interface{}{&index{0, 1}, &index{1, 3}}
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