英文:
Calling type function withouth a type
问题
我创建了一个特定类型的函数。一旦我创建了它，我可以按照预期的方式调用它，但问题是当我想在不声明函数类型的变量的情况下调用它时。
下面是一个可以澄清一切的示例：
type MyStruct struct{
number1 int
number2 int
}
func (input *MyStruct) declareValues(val1 int, val2 int){
input.number1 = val1
input.number2 = val2
}
func (input MyStruct) add() int{
return input.number1 + input.number2
}
var declared MyStruct
declared.declareValues(2,3)
fmt.Println(declared.add()) // 应该返回 5
fmt.Println(MyStruct{}.add()) // 如果成功，应该返回 0
问题是，如果我想使用更复杂的方法，并且它应该根据结构体字段是否为默认值给出不同的返回结果（因此我不需要声明变量，可以使用已声明的类型进行调用）。我必须以这种方式做，因为我不想声明一个变量来调用该方法。
英文:
I have created a function of a certain type. Once I did it, I can call it the way it's meant to be done, the problem comes when I want to call it without declaring a variable of the type of the function.
Here's an example that may clarify everything:
type MyStruct struct{
number1 int
number2 int
}
func (input *MyStruct) declareValues(val1 int, val2 int){
input.number1 = val1
input.number2 = val2
}
func (input MyStruct) add() int{
return number1 + number2
}
var declared MyStruct
declared.declareValues(2,3)
fmt.Println(declared.add()) // Should return 5
fmt.Println(¿MyStruct?.add()) // If works, should return 0
The point is that If I want to do it with a more complex method, and it should give me an answer if the fields of the struct are the default ones (so I shouldn't have to declare a variable and I could call it using the type declared) and another return if the fields are changed.
I have to do it that way because I dont want to declare a variable to call the method.
答案1
得分: 0
你可以使用MyStruct{}.add()，在playground上试一试。这仍然会分配一个方法接收器的实例，但至少你不需要将其存储在一个单独的变量中。
英文:
You can do MyStruct{}.add(), try it on playground. This still allocates an instance of the method's receiver but at least you don't have to store it in a separate variable.
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