英文:
How to make loop based on n value?
问题
假设我有一个值 n 等于 3，
代码看起来像这样：
//循环 1
for i := 0; i < len(x); i++ {
//循环 2
for j := 0; (j < len(x)) && (j != i); j++ {
//循环 3
for k := 0; (k < len(x)) && (k != i) && (k != j); k++ {
}
}
}
然而，我正在尝试弄清楚如何根据值自动生成代码，所以当 n 的值为 5 时，它应该生成：
循环 1 {
循环 2 {
循环 3 {
循环 4 {
循环 5 {
}
}
}
}
}
这可能吗？
英文:
Let say i have n value 3,
The code will look like this :
//Loop 1
for i := 0; i < len(x); i++ {
//Loop 2
for j := 0; (j < len(x)) && (j != i); j++ {
//Loop 3
for k := 0; (k < len(x)) && (k != i) && (k != j); k++ {
}
}
}
However, I was trying to figure out how to make it automatically based on the value, so that when the n value is 5, it should generate:
Loop 1 {
Loop 2 {
Loop 3 {
Loop 4 {
Loop 5 {
}
}
}
}
}
Is it possible?
答案1
得分: 1
这是一个递归的示例：https://go.dev/play/p/cLm-QHydM37
这将产生以下迭代结果：
当长度为5时：map[1:5 2:25 3:125 4:625 5:3125]
当长度为3时：map[1:3 2:9 3:27 4:0 5:0]
package main
import "fmt"
func main() {
count := map[int]int{1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
// 这个递归基于 https://gobyexample.com/recursion
var loopFunc func(current int, data []string)
loopFunc = func(current int, data []string) {
for i := 0; (i < len(data)) && (len(data) != current-1); i++ {
count[current] = count[current] + 1
loopFunc(current+1, data)
}
}
loopFunc(1, make([]string, 5))
fmt.Println(count)
}
我可能没有完全理解你的循环逻辑，但这应该可以为你提供一个起点，供你继续进行。
英文:
Here is an example of a recursion https://go.dev/play/p/cLm-QHydM37
This results in the following iterations
When length is 5: map[1:5 2:25 3:125 4:625 5:3125]
When length is 3: map[1:3 2:9 3:27 4:0 5:0]
package main
import "fmt"
func main() {
count := map[int]int{1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
// This recursion is based on https://gobyexample.com/recursion
var loopFunc func(current int, data []string)
loopFunc = func(current int, data []string) {
for i := 0; (i < len(data)) && (len(data) != current-1); i++ {
count[current] = count[current] + 1
loopFunc(current+1, data)
}
}
loopFunc(1, make([]string, 5))
fmt.Println(count)
}
I might not have your loop logic exactly right but this should be a springboard for you to continue on from.
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