英文:
How to pass integer as pointer using interface{} to function
问题
我以为这是一件简单的事情，但我错了。我不能使用interface{}将integer作为指针传递给函数。
示例：
var test int
someChange(&test)
fmt.Printf("函数之后：%d", test)
func someChange(i interface{}) error{
newFunnyValue := 42
i = newFunnyValue
fmt.Printf("来自someChange的问候，现在test的值是：%d" , i)
return nil //更改成功，返回nil
}
结果为：
来自someChange的问候，现在test的值是：42
函数之后：0
我读到interface{}类似于void*，所以上面的代码应该可以工作，但实际上不行，为什么？我想补充一下，如果我传递的是一个结构体对象，一切都正常工作。
我需要将int包装在某个结构体中吗？
编辑：
https://play.golang.org/p/rg1vabug0P
英文:
I thought this is a simple thing to do, but I was wrong. I can't pass integer as pointer to function using interface{}.
Example:
var test int
someChange(&test)
fmt.Printf("after function: %d", test)
func someChange(i interface{}) error{
newFunnyValue := 42
i = newFunnyValue
fmt.Printf("hello from someChange now value test is: %d" , i)
return nil //change gone ok, so nil
}
And result:
hello from someChange now value test is: 42
after function: 0
I read that interface{} is similar to void* so above code should work but it's not, why? I want to add that if I pass some object which is a struct, everything works good.
Do I have to wrap int in some struct?
Edit:
答案1
得分: 3
如果你想观察someChange()函数外部的变化（在test变量中），你必须修改指向的值（给它赋一个新值）。你没有这样做，你只是给i参数赋了一个新值（它是someChange()函数内部的局部变量）。
你可以使用类型断言从i接口变量中获取*int指针，然后可以给指向的值赋一个新值。
示例：
func someChange(i interface{}) error {
newFunnyValue := 42
if p, ok := i.(*int); ok {
*p = newFunnyValue
return nil //change gone ok, so nil
}
return errors.New("Not *int")
}
测试：
var test int
someChange(&test)
log.Printf("after function: %d", test)
输出（在Go Playground上尝试）：
2009/11/10 23:00:00 after function: 42
请注意，将int值或*int指针包装在结构体中是不必要的，如果你没有给指向的对象赋一个新值，它不会有任何区别。
英文:
If you want to observe the change outside of the someChange() function (in the test variable), you must modify the pointed value (assign a new value to it). You're not doing that, you just assign a new value to the i parameter (which is a local variable inside someChange()).
You may obtain the *int pointer from the i interface variable using type assertion, and then you can assign a new value to the pointed value.
Example:
func someChange(i interface{}) error {
newFunnyValue := 42
if p, ok := i.(*int); ok {
*p = newFunnyValue
return nil //change gone ok, so nil
}
return errors.New("Not *int")
}
Testing it:
var test int
someChange(&test)
log.Printf("after function: %d", test)
Output (try it on the Go Playground):
2009/11/10 23:00:00 after function: 42
Note that wrapping the int value or the *int pointer in a struct is unnecessary and it wouldn't make a difference if you're not assigning a new value to the pointed object.
答案2
得分: 1
i在func someChange()中仍然是interface{}类型。在赋值过程中，你需要对其进行类型转换：
*i.(*int) = 42
英文:
i is still of type interface{} within the func someChange(). You have to cast it during the assignment:
*i.(*int) = 42
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