英文:
Is there a shortcut for assigning a variable to a pointer without creating the variable in a separate line first?
问题
如果我有这样一个结构体：
type Message struct {
Id int64
Message string
ReplyTo *int64
}
然后如果我像这样创建一个结构体的实例：
var m Message
m.Id = 1
m.Message = "foo bar yo"
var replyTo = int64(64)
m.ReplyTo = &replyTo
那么它会起作用。
但我想知道是否有一个快捷方式来完成最后一步？
我尝试了这样的操作：
m.ReplyTo = &int64{64}
但它不起作用。
英文:
If I have a struct like this:
type Message struct {
Id int64
Message string
ReplyTo *int64
}
And then if I did create an instance of this struct like this:
var m Message
m.Id = 1
m.Message = "foo bar yo"
var replyTo = int64(64)
m.ReplyTo = &replyTo
Then it would work.
But I was wondering if there was a shortcut for the last step?
I tried doing something like:
m.ReplyTo = &int64{64}
But it did not work.
答案1
得分: 2
我不认为你可以这样做，因为该值是一个原始值，尝试像下面这样一次性操作会导致语法错误。它试图获取一个值的地址，所以这是不可能的。至少我不知道有一种可以实现的方式。
someInt := &int64(10) // 不会编译通过
你还有另一种选择，可以编写一个函数来返回指向原始值的指针，如下所示：
func NewIntPointer(value int) *int {
return &value
}
英文:
I don't think you can because the value is a primitive and attempting to do it in one shot like the below would be a syntax error. Its attempting to get an address of a value so it wouldn't be possible. At least I am not aware of a way where its possible.
someInt := &int64(10) // would not compile
The other alternative you have is to write a function to return a pointer to the primitive like the following:
func NewIntPointer(value int) *int {
return &value
}
答案2
得分: 1
获取int指针的一个巧妙方法，而不需要创建新的变量。
someIntPtr := &[]int64{10}[0]
英文:
A tricky way to get int pointer without create new variable.
someIntPtr := &[]int64{10}[0]
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