英文:
Go unpacking array as arguments to path.Join
问题
我想解开字符串数组并传递给path.Join函数。
package main
import (
"fmt"
"path"
)
func main() {
p := []string{"a", "b", "c"}
fmt.Println(path.Join(p...))
}
这段代码的输出结果是：
a/b/c
但是如果我像这样传递参数：
package main
import (
"fmt"
"path"
)
func main() {
p := []string{"a", "b", "c"}
fmt.Println(path.Join("d", p...))
}
它不起作用。
tmp/sandbox299218161/main.go:10: too many arguments in call to path.Join
have (string, []string...)
want (...string)
我认为我对解包有误解，有什么建议吗？
英文:
I want to unpack string array and pass to path.Join
package main
import (
"fmt"
"path"
)
func main() {
p := []string{"a", "b", "c"}
fmt.Println(path.Join(p...))
}
Output of this code is:
a/b/c
But if I pass arguments like:
package main
import (
"fmt"
"path"
)
func main() {
p := []string{"a", "b", "c"}
fmt.Println(path.Join("d", p...))
}
It doesn't work.
tmp/sandbox299218161/main.go:10: too many arguments in call to path.Join
have (string, []string...)
want (...string)
I think I have misunderstanding on unpacking, any suggestions?
答案1
得分: 10
你没有真正误解任何事情，你只是不能这样做。规范中说：
> 如果最后一个参数可以赋值给切片类型[]T，并且在参数后面跟着...，则它可以作为...T参数的值传递，而不需要创建新的切片。
简而言之，p...只能作为参数的整个可变部分使用，因为当你这样做时，它只是在函数内部将p作为参数切片重复使用，而不是创建一个新的切片。
如果你想在开头添加一些参数，你需要先构建一个包含所有参数的切片，类似于：
p := []string{"a", "b", "c"}
p2 := append([]string{"d"}, p...)
fmt.Println(path.Join(p2...))
这样可以正常工作，并打印出"d/a/b/c"。
英文:
You're not misunderstanding anything really, you just can't do that. The spec says:
> If the final argument is assignable to a slice type []T, it may be passed unchanged as the value for a ...T parameter if the argument is followed by .... In this case no new slice is created.
In short, p... can only be used as the entire variadic part of the arguments, because when you do that, it simply re-uses p as the parameter slice within the function, instead of making a new one.
If you want to add some arguments at the beginning, you would have to construct your own slice with all of the arguments first, something like:
p := []string{"a", "b", "c"}
p2 := append([]string{"d"}, p...)
fmt.Println(path.Join(p2...))
which works alright and prints "d/a/b/c".
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