英文:
How to "cast" a pointer back to a value in Golang?
问题
我正在使用time.Time作为我的结构体之一中的指针。例如：
type struct Example{
CreatedDate *time.Time
}
我使用指针，这样如果没有日期存在，我就可以将nil传递给结构体。
然而，现在出现了一个问题，因为我需要使用time.Since(then)函数，该函数不接受指针作为参数，而是接受time.Time。
所以...
在结构体前面加上"&"很容易，例如&time.Time，但如果你有一个指针，你如何将其转换回time.Time类型呢？
例如（不起作用，但可能会给你一个想法）：
var t *time.Time = &time.Now()
var t2 time.Time = t.(time.Time)
我希望你能帮助我。这个问题感觉很愚蠢，因为当我在谷歌上搜索时找不到任何相关信息。我感觉我在这里只是漏掉了什么东西;-)
英文:
I'm using time.Time as a pointer in one of my structs. Eg
type struct Example{
CreatedDate *time.Time
}
I'm using the pointer, so I can pass nil to the struct, if no date exists.
This does however pose a problem now, since I need to use the time.Since(then) function which does not accept the pointer as the parameter, but takes a time.Time instead.
So...
It's quite easy to put "&" in front of a struct eg. &time.Time, but if you have a pointer, how can you reverse it, back into eg. a type of time.Time?
Eg. (does not work, but might give you an idea of what I mean)
var t *time.Time = &time.Now()
var t2 time.Time = t.(time.Time)
I hope you can help. The question feels quite silly, since I can't find anything about it when googling. I get the feeling I'm just missing something here 
答案1
得分: 89
var t *time.Time = &time.Now()
var t2 time.Time = *t
就像在C语言中一样，&表示“取地址”，并将其分配给一个标记为*T的指针。但是*也表示“取值”，或者解引用指针，并将值分配给变量。
这有点离题，但我认为*的这种双重用途部分是让C语言新手对指针感到困惑的原因之一。
英文:
var t *time.Time = &time.Now()
var t2 time.Time = *t
Just like in C, & means "address of" and assigns a pointer which is marked as *T. But * also means "value of", or dereferencing the pointer, and assigns the value.
This is a bit off-topic, but IMO this dual use of * is partly what gets newbies in C and friends confused about pointers.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论