douyuan1049 2017-08-02 18:08
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切片结构与指向结构的指针

I don't understand the behavior of the following piece of code. In creating a list of matching structs as a slice of struct pointers, the code always prints the last element of original array (which actually wasn't a match)—it prints 12 and 12. However, if I change matches to be []Widget instead of []*Widget, then it will print 10 and 11.

Why is this?

package main

import (
    "fmt"
)

func main() {

    type Widget struct {
        id    int
        attrs []string
    }

    widgets := []Widget{
        Widget{
            id:    10,
            attrs: []string{"blah", "foo"},
        },
        Widget{
            id:    11,
            attrs: []string{"foo", "bar"},
        },
        Widget{
            id:    12,
            attrs: []string{"xyz"},
        },
    }

    var matches []*Widget
    for _, w := range widgets {
        for _, a := range w.attrs {
            if a == "foo" {
                matches = append(matches, &w)
                break
            }
        }
    }

    for _, m := range matches {
        fmt.Println(m.id)
    }

}
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1条回答 默认 最新

  • dtzhfb4869 2017-08-02 18:18
    关注

    That's because when you use the pointers you are adding &w to the array.

    Note that w is actually the local variable used in the loop, so that's not the address you want to add to the matches array.

    (even though the value of the variable w changes through the loop, it's address stays the same)

    When the loop ends, w ends up with the last value so that's why it prints 12 two times.

    You need to add the address of the element that matched instead.

    If you do this:

    matches = append(matches, &widgets[i])
    

    Then it'd work fine with pointers as well.

    Modified Go playground for you to test it:

    https://play.golang.org/p/YE-cokyEHu

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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