英文:
Go - append to slice in struct
问题
我正在尝试实现以下两个简单的结构体：
package mainimport ("fmt")type MyBoxItem struct {Name string}type MyBox struct {Items []MyBoxItem}func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {return append(box.Items, item)}func main() {item1 := MyBoxItem{Name: "Test Item 1"}item2 := MyBoxItem{Name: "Test Item 2"}items := []MyBoxItem{}box := MyBox{items}AddItem(box, item1) // 这是我卡住的地方fmt.Println(len(box.Items))}
我做错了什么？我只是想在box结构体上调用addItem方法并传入一个item。
英文:
I am trying to implement 2 simple structs as follows:
package mainimport ("fmt")type MyBoxItem struct {Name string}type MyBox struct {Items []MyBoxItem}func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {return append(box.Items, item)}func main() {item1 := MyBoxItem{Name: "Test Item 1"}item2 := MyBoxItem{Name: "Test Item 2"}items := []MyBoxItem{}box := MyBox{items}AddItem(box, item1) // This is where i am stuckfmt.Println(len(box.Items))}
What am i doing wrong? I simply want to call the addItem method on the box struct and pass an item in
答案1
得分: 108
嗯... 这是在Go中向切片添加元素时人们最常犯的错误。你必须将结果重新赋值给切片。
func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {box.Items = append(box.Items, item)return box.Items}
另外，你已经为*MyBox类型定义了AddItem方法，所以调用这个方法应该是box.AddItem(item1)。
英文:
Hmm... This is the most common mistake that people make when appending to slices in Go. You must assign the result back to slice.
func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {box.Items = append(box.Items, item)return box.Items}
Also, you have defined AddItem for *MyBox type, so call this method as box.AddItem(item1)
答案2
得分: 21
package main
import (
"fmt"
)
type MyBoxItem struct {
Name string
}
type MyBox struct {
Items []MyBoxItem
}
func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {
box.Items = append(box.Items, item)
return box.Items
}
func main() {
item1 := MyBoxItem{Name: "Test Item 1"}items := []MyBoxItem{}box := MyBox{items}box.AddItem(item1)fmt.Println(len(box.Items))
}
英文:
package mainimport ("fmt")type MyBoxItem struct {Name string}type MyBox struct {Items []MyBoxItem}func (box *MyBox) AddItem(item MyBoxItem) []MyBoxItem {box.Items = append(box.Items, item)return box.Items}func main() {item1 := MyBoxItem{Name: "Test Item 1"}items := []MyBoxItem{}box := MyBox{items}box.AddItem(item1)fmt.Println(len(box.Items))}
Output:
1
答案3
得分: 17
虽然两种答案都是完全正确的。还有两个更改可以进行：
- 摆脱返回语句，因为该方法是为指向结构体的指针调用的，所以切片会自动修改。
- 没有必要初始化一个空切片并将其分配给结构体。
package mainimport ("fmt")type MyBoxItem struct {Name string}type MyBox struct {Items []MyBoxItem}func (box *MyBox) AddItem(item MyBoxItem) {box.Items = append(box.Items, item)}func main() {item1 := MyBoxItem{Name: "Test Item 1"}item2 := MyBoxItem{Name: "Test Item 2"}box := MyBox{}box.AddItem(item1)box.AddItem(item2)// checking the outputfmt.Println(len(box.Items))fmt.Println(box.Items)}
英文:
Though both answers are perfectly fine. There are two more changes that can be done,
- Getting rid of the return statement as the method is called for a pointer to the struct, so the slice is automatically modified.
- There is no need to initialise an empty slice and assign it to the struct
package mainimport ("fmt")type MyBoxItem struct {Name string}type MyBox struct {Items []MyBoxItem}func (box *MyBox) AddItem(item MyBoxItem) {box.Items = append(box.Items, item)}func main() {item1 := MyBoxItem{Name: "Test Item 1"}item2 := MyBoxItem{Name: "Test Item 2"}box := MyBox{}box.AddItem(item1)box.AddItem(item2)// checking the outputfmt.Println(len(box.Items))fmt.Println(box.Items)}
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