英文:
Generate array with with range of integers
问题
我有两个值：
> [3:6]
我试图在Golang中尝试一些操作，但是我找不到一个好的方法来根据这些值创建一个数组。
这是我想要实现的：
[3,4,5,6]
英文:
I have two values:
> [3:6]
I was trying to play something around in Golang but I can't manage to find a good method to create an array according to those values.
This what I would like to achieve:
[3,4,5,6]
答案1
得分: 5
你可以使用for ... range结构使代码更简洁，甚至可能更快：
lo, hi := 3, 6s := make([]int, hi-lo+1)for i := range s {s[i] = i + lo}
作为一种好奇，循环可以在没有循环变量的情况下实现，但速度会较慢，代码也会更长。通过递减hi：
for ; hi >= lo; hi-- {s[hi-len(s)+1] = hi}
或者递增lo：
for ; lo <= hi; lo++ {s[len(s)-1-hi+lo] = lo}
英文:
You can make use of the for ... range construct to make it more compact and maybe even faster:
lo, hi := 3, 6s := make([]int, hi-lo+1)for i := range s {s[i] = i + lo}
As a matter of curiosity, the loop can be implemented without a loop variable, but it will be slower, and the code longer. By decrementing hi:
for ; hi >= lo; hi-- {s[hi-len(s)+1] = hi}
Or incrementing lo:
for ; lo <= hi; lo++ {s[len(s)-1-hi+lo] = lo}
答案2
得分: 0
我正在寻找一个类似的答案，@icza的答案是一个很好的起点。最终，我根据icza的答案创建了一个辅助函数：
func createNumbers(lo int, hi int) []int {s := make([]int, hi-lo+1)for i:= range s {s[i] = i + lo}return s}
英文:
I was looking for a similar answer and @icza's answer was a great start. I ended up creating a helper function based on icza's answer:
func createNumbers(lo int, hi int) []int {s := make([]int, hi-lo+1)for i:= range s {s[i] = i + lo}return s}
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