Golang reflect反射调用函数

package main

import (
"fmt"
"reflect"
)
type Foo struct {
    One string
    Two int
}

func (f Foo) CallFuncNoArgs() {
    fmt.Printf("One is: %s, Two is: %d\n", f.One, f.Two)

}
func (f Foo) CallFuncWithArgs(name string, a int) {
    fmt.Printf("name:%s a:%d\n", name, a)
}

func (f Foo) CallFuncWithArgsAndRet(one, two int) (int, string) {
    fmt.Printf("one:%d two:%d\n", one, two)
    return one + two, "CallFuncWithArgsAndRet string"
}

func main() {
    var foo = &Foo{"Award", 666}
    // 方法名必须大写
    // if name not FuncName will panic  reflect: call of reflect.Value.Call on zero Value
    reflect.ValueOf(foo).MethodByName("CallFuncNoArgs").Call([]reflect.Value{})

    in := make([]reflect.Value, 2)
    in[0] = reflect.ValueOf("Colin")
    in[1] = reflect.ValueOf(111)
    _ = reflect.ValueOf(foo).MethodByName("CallFuncWithArgs").Call(in)

    in[0] = reflect.ValueOf(1)
    in[1] = reflect.ValueOf(2)
    values := reflect.ValueOf(foo).MethodByName("CallFuncWithArgsAndRet").Call(in)
    //fmt.Printf("return value:%v\n", value[0].Interface().(int))
    for _,value := range values {
        fmt.Printf("value:%v\n", value)
    }
}

1.CallFuncNoArgs是无参数无返回值调用
2.CallFuncWithArgs是有参数无返回值调用
3.CallFuncWithArgsAndRet是有参数有返回值调用
重点说明：Call(in []reflect.ValueOf),必须传入Value类型的切片

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