英文:
Is a slice's underlying array accessible if the slice is shrunk?
问题
给定一个类型，如下所示：
type LicenseCards struct {
cards *[]int
}
我不会展示创建切片的代码。但是这段代码会移除顶部的元素，忽略长度为零的情况。
func (licenseCards *LicenseCards) PopLicenseCard() int {
l := len(*licenseCards.cards)
ret := (*licenseCards.cards)[l-1]
*licenseCards.cards = (*licenseCards.cards)[:l-1]
return ret
}
如果我从切片中移除最后一个元素并返回指向被移除元素的指针，那么它是否保证仍然可用？
英文:
Given a type such as:
type LicenseCards struct {
cards *[]int
}
I won't show the code that creates the slice. But this removes the top item, ignoring the zero-length case.
func (licenseCards *LicenseCards) PopLicenseCard() int {
l := len(*licenseCards.cards)
ret := (*licenseCards.cards)[l-1]
*licenseCards.cards = (*licenseCards.cards)[:l-1]
return ret
}
If I remove the last item from the slice and return a pointer to the removed item, is it guaranteed to still be available?
答案1
得分: 2
正如@Volker所说，如果有东西在使用内存，垃圾回收器将不会释放它。
你的代码中还有一个问题，就是在使用.运算符之前不需要解引用指针（使用*运算符），只需这样做：l := len(licenseCards.cards)。
此外，你不需要将cards和接收器都设置为指针。如果你不介意的话，我想建议这样做：
type LicenseCards struct {
cards []int
}
func (lc *LicenseCards) PopLicenseCard() int {
l := len(lc.cards)
ret := lc.cards[l-1]
lc.cards = lc.cards[:l-1]
return ret
}
英文:
As @Volker said the memory will not be released by the GC if something is using it.
Another point with your code is that you do not need to dereference a pointer (using * operator) before using the . operator eg: just do this: l := len(licenseCards.cards).
Also you don't need cards and the receiver to both be pointers. If you don't mind I would like to suggest this:
type LicenseCards struct {
cards []int
}
func (lc *LicenseCards) PopLicenseCard() int {
l := len(lc.cards)
ret := lc.cards[l-1]
lc.cards = lc.cards[:l-1]
return ret
}
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