Question

例如，我有两个不同的结构Foo和Bar

struct Foo {...}
struct Bar {...}

并且在它们上构建了许多函数和其他类型。

C ++风味：

template<typename T>
struct Identified {
    T model;
    std::string id;
};

template<typename T>
Identified<T> GetIdentifiedModel(std::string id) {
    Identified<T> result;
    T.id = id;
    T.model.set(getSomeData(id));  // Common method for T
    return result;
}

如何在Go中实现这些示例？

对于常用方法，接口可以完成这项工作，但是我没有看到如何从接口中检索特定类型来声明它，返回它或其他任何东西，我再也无法处理复制/粘贴代码了：）

谢谢！

编辑@Amd：https://ideone.com/rqpsQb

#include <string>
#include <iostream>

struct Foo {
    char c;
    void set(std::string s) {c = s[0];}; // We don't really care here
};
struct Bar {
    int n;
    void set(std::string s) {n = s.size();}; // We don't really care here
};

template<typename T>
struct Identified {
    T model;
    std::string id;
};

template<typename T>
Identified<T> GetIdentifiedModel(std::string id) {
    Identified<T> result;
    result.id = id;
    // Obviously shouldn't be ID but for the example
    result.model.set(id);  // Common method for T
    return result;
}

void assert(bool b) {
    if (b) std::cout << "OK" << std::endl;
    else std::cout << "There is a problem !" << std::endl;
};

int main() {
    auto fooWithID = GetIdentifiedModel<Foo>("foo id");
    auto barWithID = GetIdentifiedModel<Bar>("bar");
    assert (fooWithID.model.c == 'f');
    assert (barWithID.model.n == 3);
    return (0);
}

Answer 1

1-您可以使用

fooWithID := GetIdentifiedModel("foo id", &Foo{})

喜欢这个工作样本（试试The Go Playground）：

package main

import "fmt"

type Foo struct {
    c byte
}

func (t *Foo) set(s string) { t.c = s[0] }

type Bar struct {
    n int
}

func (t *Bar) set(s string) { t.n = len(s) }

type Identified struct {
    model T
    id    string
}

func GetIdentifiedModel(id string, t T) *Identified {
    result := &Identified{model: t}
    result.id = id
    result.model.set(id)
    return result
}

func assert(b bool) {
    if b {
        fmt.Println("OK")
    } else {
        fmt.Println("There is a problem !")
    }
}

func main() {
    fooWithID := GetIdentifiedModel("foo id", &Foo{})
    barWithID := GetIdentifiedModel("bar", &Bar{})

    assert(fooWithID.model.(*Foo).c == 'f')
    assert(barWithID.model.(*Bar).n == 3)
}

type T interface {
    set(string)
}

输出：

OK
OK

2-您可以使用（这很好看：Identified model Foo）：

fooWithID := GetIdentifiedModel("foo id", &Identified{model: &Foo{}})

喜欢这个工作样本（试试The Go Playground）：

package main

import "fmt"

type Foo struct {
    c byte
}

func (t *Foo) set(s string) { t.c = s[0] }

type Bar struct {
    n int
}

func (t *Bar) set(s string) { t.n = len(s) }

type Identified struct {
    model T
    id    string
}

func GetIdentifiedModel(id string, result *Identified) *Identified {
    result.id = id
    result.model.set(id)
    return result
}

func assert(b bool) {
    if b {
        fmt.Println("OK")
    } else {
        fmt.Println("There is a problem !")
    }
}

func main() {
    fooWithID := GetIdentifiedModel("foo id", &Identified{model: &Foo{}})
    barWithID := GetIdentifiedModel("bar", &Identified{model: &Bar{}})    
    assert(fooWithID.model.(*Foo).c == 'f')
    assert(barWithID.model.(*Bar).n == 3)
}

type T interface {
    set(string)
}

输出：

OK
OK

另请参阅：One method to handle all the struct types that embed one common struct (json marshalling)

Go中的C ++类似模板的行为

1 个答案: