英文:
Go: Define multidimensional array with existing array's type and values?
问题
可以使用现有数组来定义和初始化一个新的多维数组吗？就像下面的代码中，而不是使用<code>var b [2][3]int</code>，只是像<code>var b [2]a</code>这样说？使用a的类型，无论它是什么，而不是硬编码它（这样会错过使用[...]的意义）。同时处理初始化=复制值的问题？
package main
func main () {
a := [...]int{4,5,6}
var b [2][3]int
b[0],b[1] = a,a
}
（我知道切片的简便性和方便性，但这个问题是关于理解数组的。）
编辑：我不敢相信我忘记了<code>var b [2][len(a)]int</code>，初学者的脑袋冻结。一行答案是<code>var b = [2][len(a)]int{a,a}</code>。那是一种类型转换，对吗？
英文:
Is it possible to a)define b)initialize a new multidimensional array using an existing array, like in following code instead of <code>var b [2][3]int</code>, just saying something like <code>var b [2]a</code> ?<br>
Using a's type whatever it is, instead of hardcoding it (which misses the point of using [...] for a).<br>
And perhaps handling initialization=copying of values at the same time?
package main
func main () {
a := [...]int{4,5,6}
var b [2][3]int
b[0],b[1] = a,a
}
(I'm aware of ease and convenience of slices, but this question is about understanding arrays.)
Edit: can't believe I forgot about <code>var b [2][len(a)]int</code>, beginner's brain freeze. One line answer would be <code>var b = [2][len(a)]int{a,a}</code> . That's a type conversion, right?
答案1
得分: 5
以下代码也可以工作。你的例子和我的例子都做了同样的事情，两者之间的速度应该没有太大差异。
除非你使用反射(reflect)来创建一个你的[3]int的切片(slice)（而不是数组），否则在你的新类型中不重复[3]int是不可能的。即使在当前版本中也不可能。在Go 1.1中将会发布。
package main
import "fmt"
func main() {
a := [...]int{4,5,6}
var b = [2][3]int{a, a}
fmt.Println(b)
}
英文:
The following code would also work. Both your example and mine do the same thing and neither should be much faster than the other.
Unless you use reflect to make a slice (not array) of your [3]int, it is impossible to not repeat [3]int in your new type. Even that is not possible in the current release. It is in tip and will be released in Go 1.1.
package main
import "fmt"
func main() {
a := [...]int{4,5,6}
var b = [2][3]int{a, a}
fmt.Println(b)
}
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