例如我有两个不同的结构 Foo
和 Bar
struct Foo {...}
struct Bar {...}
以及基于它们构建的许多函数和其他类型。
C++ 风格:
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(interface)可以完成工作,但我看不到如何从接口(interface)检索特定类型来声明它、返回它或其他任何东西,我无法再处理复制/粘贴代码:)
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);
}
最佳答案
1- 你可以使用
fooWithID := GetIdentifiedModel("foo id", &Foo{})
喜欢这个工作示例(尝试 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- 你可以使用(读起来很舒服:Identified model Foo
):
fooWithID := GetIdentifiedModel("foo id", &Identified{model: &Foo{}})
喜欢这个工作示例(尝试 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)
关于Go 中类似 C++ 模板的行为,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39559293/