在赋值时解包切片吗？

huangapple
go评论116阅读模式
英文:

Unpack slices on assignment?

问题

在Go语言中，有没有一种像Python中那样从数组中进行多重赋值的优雅方式？以下是我尝试做的一个Python示例（将一个字符串拆分并将结果数组分配给两个变量）。

  1. python:
  2. >>> a, b = "foo;bar".split(";")

我目前的解决方案是：

  1. x := strings.Split("foo;bar", ";")
  2. a, b := x[0], x[1]

我可以看到在某些情况下，这种方法可能会变得混乱。我目前面临的实际例子是解析书签文件并将其分配给一个映射：

  1. bookmark := make(map[string]string)
  2. x := strings.Split("foo\thttps://bar", "\t")
  3. name, link := x[0], x[1]
  4. bookmark[name] = link

现在我有一个无用的变量x。我想做的是：

  1. bookmark := make(map[string]string)
  2. name, link := strings.Split("foo\thttps://bar", "\t")
  3. bookmark[name] = link

但这是无效的。

英文:

Is there an elegant way in Go to do multiple assignments from arrays like in Python? Here is a Python example of what I'm trying to do (split a string and then assign the resulting array into two variables).

  1. python:
  2. >>> a, b = "foo;bar".split(";")

My current solution is:

  1. x := strings.Split("foo;bar", ";")
  2. a, b := x[0], x[1]

I'm can see this getting messy in some constructs. The practical example I'm currently facing is a bookmark file parsing and assigning to a map:

  1. bookmark := make(map[string]string)
  2. x := strings.Split("foo\thttps://bar", "\t")
  3. name, link := x[0], x[1]
  4. bookmark[name] = link

Now I have a useless variable x sitting around. I'd like to do something like:

  1. bookmark := make(map[string]string)
  2. name, line := strings.Split("foo\thttps://bar", "\t")
  3. bookmark[name] = link

but that's invalid.

答案1

得分: 69

如Sergio Tulentsev所提到的，Python中的通用打包/解包操作在这里是不支持的。我认为解决这个问题的方法是定义一个自己的小型临时函数，使用多个返回值：

  1. func splitLink(s, sep string) (string, string) {
  2.     x := strings.Split(s, sep)
  3.     return x[0], x[1]
  4. }

然后你可以这样写：

  1. name, link := splitLink("foo\thttps://bar", "\t")

但是这显然只适用于至少拆分两个子字符串的情况，并且如果拆分的子字符串超过两个，它会默默地忽略掉。如果你经常使用这种方式，它可能会使你的代码更易读。

--编辑--

另一种解包数组的方法是使用可变指针参数：

  1. func unpack(s []string, vars... *string) {
  2.     for i, str := range s {
  3.         *vars[i] = str
  4.     }
  5. }

这样你可以这样写：

  1. var name, link string
  2. unpack(strings.Split("foo\thttps://bar", "\t"), &name, &link)
  3. bookmarks[name] = link

这种方法适用于任何数组大小，但可读性较差，并且你必须显式声明变量。

英文:

As Sergio Tulentsev mentioned, general packing/unpacking as is done in Python is not supported. I think the way to go there is to define your own small ad-hoc function using multiple return values:

  1. func splitLink(s, sep string) (string, string) {
  2.     x := strings.Split(s, sep)
  3.     return x[0], x[1]
  4. }

And you can then write:

  1. name, link := splitLink("foo\thttps://bar", "\t")

But this will obviously work only when at least two substrings are being split, and silently ignore if more than two were. If this is something you use a lot, it might make your code more readable though.

--EDIT--

Another way to unpack an array is via variadic pointer arguments:

  1. func unpack(s []string, vars... *string) {
  2.     for i, str := range s {
  3.         *vars[i] = str
  4.     }
  5. }

Which let you write:

  1. var name, link string
  2. unpack(strings.Split("foo\thttps://bar", "\t"), &name, &link)
  3. bookmarks[name] = link

This will work for any array size, but it is arguably less readable, and you have to declare your variables explicitly.

答案2

得分: 16

你也可以使用匿名函数：

  1. a, b := func() (string, string) {
  2.     x := strings.Split("foo;bar", ";")
  3.     return x[0], x[1]
  4. }()
  6. 注意：不要忘记在闭合括号 `}` 的末尾加上 `()`，否则会出现错误：
  8. ```assignment mismatch: 2 variable but 1 values```
  10. 这是因为没有加上 `()`，返回的是一个函数（1个值），而不是预期的字符串（2个值）。
  12. <details>
  13. <summary>英文:</summary>
  15. You could also use anonymous functions:
  17.     a, b := func() (string, string) {
  18.         x := strings.Split(&quot;foo;bar&quot;, &quot;;&quot;)
  19.         return x[0], x[1]
  20.     }()
  22. Note: don&#39;t forget the `()` on the end of the closing bracket `}` otherwise you will get the error:
  24. ```assignment mismatch: 2 variable but 1 values```
  26. This is because without the `()` a function (1 value) is returned not the expected strings (2 values).
  30. </details>
  34. # 答案3
  35. **得分**: 9
  37. 如果你的函数的目的是仅通过分隔符的第一个出现来拆分字符串，你可以自己编写一个函数：
  39. ```go
  40. package main
  42. import (
  43. 	"fmt"
  44. 	"strings"
  45. )
  47. func Split(s, sep string) (string, string) {
  48. 	// 空字符串应该返回空
  49. 	if len(s) == 0 {
  50. 		return s, s
  51. 	}	
  53. 	slice := strings.SplitN(s, sep, 2)
  55. 	// 如果没有分隔符
  56. 	if len(slice) == 1 {
  57. 		return slice[0], ""
  58. 	}
  59. 		
  60. 	return slice[0], slice[1]
  61. }
  63. func main() {
  64. 	a, b := Split("foo;bar;foo", ";")
  65. 	fmt.Println(a, b)
  66. }

输出：

foo bar;foo

Playground

英文:

If your function is meant to split a string only by the first occurrence of the separator, you can always make your own function:

  1. package main
  2. import (
  3. &quot;fmt&quot;
  4. &quot;strings&quot;
  5. )
  6. func Split(s, sep string) (string, string) {
  7. // Empty string should just return empty
  8. if len(s) == 0 {
  9. return s, s
  10. }	
  11. slice := strings.SplitN(s, sep, 2)
  12. // Incase no separator was present
  13. if len(slice) == 1 {
  14. return slice[0], &quot;&quot;
  15. }
  16. return slice[0], slice[1]
  17. }
  18. func main() {
  19. a, b := Split(&quot;foo;bar;foo&quot;, &quot;;&quot;)
  20. fmt.Println(a, b)
  21. }

Output:

>foo bar;foo

Playground

huangapple
  • 本文由 发表于 2013年11月7日 17:27:32
  • 转载请务必保留本文链接：https://go.coder-hub.com/19832189.html
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