一个通用函数，用于设置不同结构字段作为映射值

拥有具有共同字段的结构...

type definition struct {
        id string
        ...
}
type requirement struct {
        id string
        ...
}
type campaign struct {
        id string
        ...
}

...我有多个这样的函数：

func filldefinitionids(values *map[string]definition) {           
        for key, value:=range *values { // repeated code
                value.id=key            // repeated code
                (*values)[key]=value    // repeated code
        }                               // repeated code
}
func fillrequirementids(values *map[string]requirement) {           
        for key, value:=range *values { // repeated code
                value.id=key            // repeated code
                (*values)[key]=value    // repeated code
        }                               // repeated code
}
func fillcampaignids(values *map[string]campaign) {           
        for key, value:=range *values { // repeated code
                value.id=key            // repeated code
                (*values)[key]=value    // repeated code
        }                               // repeated code
}

我想要一个单一的函数，用泛型（或接口，等等）来概括访问，有点......

func fillIds[T Definition|Requirement|Campaign](values *map[string]T) {           
        for key, value:=range *values {
                value.Id=key
                (*values)[key]=value
        }                                
}

当然，这会导致 value.id 未定义（类型 t 没有字段或方法 id）。我已经多次能够克服类似的问题，但这次我找不到解决方案。

如何将这组函数抽象为一个函数？

正确答案

type definition struct {
    id string
}
type requirement struct {
    id string
}
type campaign struct {
    id string
}

func (v definition) withid(id string) definition   { v.id = id; return v }
func (v requirement) withid(id string) requirement { v.id = id; return v }
func (v campaign) withid(id string) campaign       { v.id = id; return v }

type withid[t any] interface {
    withid(id string) t
}

func fillids[t withid[t]](values map[string]t) {
    for key, value := range values {
        values[key] = value.withid(key)
    }
}

func main() {
    m1 := map[string]definition{"foo": {}, "bar": {}}
    fillids(m1)
    fmt.println(m1)

    m2 := map[string]campaign{"abc": {}, "def": {}}
    fillids(m2)
    fmt.println(m2)
}

https://go.dev/play/p/f3qk0gcykea

如果需要使用值映射，则可以替代@blackgreen的答案。

type common struct {
    id string
}

func (v *common) setid(id string) { v.id = id }

type definition struct {
    common
}
type requirement struct {
    common
}
type campaign struct {
    common
}

type idsetter[t any] interface {
    *t
    setid(id string)
}

func fillids[t any, u idsetter[t]](values map[string]t) {
    for key, value := range values {
        u(&value).setid(key)
        values[key] = value
    }
}

func main() {
    m1 := map[string]definition{"foo": {}, "bar": {}}
    fillids(m1)
    fmt.println(m1)

    m2 := map[string]campaign{"abc": {}, "def": {}}
    fillids(m2)
    fmt.println(m2)
}

https://go.dev/play/p/ag050b0pefw