英文:
How to define generic function with custom structs without listing all of them?
问题
假设我有两个不同的结构体：
type One struct {Id string// 其他字段}type Two struct {Id string// 其他字段}
是否有可能定义一个函数，可以接受One和Two，而不需要显式地列出它们作为选项？
例如，我想要这样的函数：
type ModelWithId struct {Id string}func Test[M ModelWithId](m M) {fmt.PrintLn(m.Id)}one := One { Id: "1" }Test(one) // 输出 1
我不想使用func Test[M One | Two](m M)，因为我可能会有10个以上的结构体，我不想每次在代码库中添加新的结构体时都要回到函数中去修改。
英文:
Let's say I have two different structs:
type One struct {Id string// Other fields}type Two struct {Id string// Other fields}
Is it possible to define a function that accepts both One and Two without explicitly listing them as options?
E.g. I am looking for something like this:
type ModelWithId struct {Id string}func Test[M ModelWithId](m M) {fmt.PrintLn(m.Id)}one := One { Id: "1" }Test(one) // Prints 1
I don't want to use funcTest[M One | Two](m M), because I'll likely have 10+ structs and I don't want to come back to the function every time I add a new struct to the codebase.
答案1
得分: 3
泛型约束使用方法来限制类型参数的行为，因此您需要将代码重写为：
type One struct {id string}func (o *One) Id() string {return o.id}然后您的使用场景将变为：type ModelWithId interface {Id() string}func Test[M ModelWithId](m M) {fmt.Println(m.Id())}
英文:
Generics constraints the type parameter behaviours using methods, so you need to rewrite your code as:
type One struct {id string}func (o *One) Id() string {return o.id}
then your use site would become:
type ModelWithId interface {Id() string}func Test[M ModelWithId](m M) {fmt.Println(m.Id())}
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