英文:
map with function pointer as key in go
问题
我最近在我的golang项目中使用了一个包含函数指针作为键的映射,就像这样:
map[*functiontype] somestructtype
我的一个同事说这是一个不好的主意,所以现在我对这个是否可行感到不确定。起初我认为这是可以的,因为方法指针可以进行相等性检查并且是不可变的。有人能提供一些关于这个问题的理由吗?
完整的示例代码如下:
package main
import "fmt"
type s struct {
string
}
type f func() string
func func1() string { return "func 1" }
func func2() string { return "func 2" }
func main() {
// 创建两个函数和两个指向它们的指针
f1, f2 := func1, func2
p1, p2 := (*f)(&f1), (*f)(&f2)
// 创建一个函数指针的映射
m := make(map[*f]s)
m[p1] = s{"struct 1"}
m[p2] = s{"struct 2"}
// 打印映射
printmapping(m, p1, p2)
// 反转指针并打印
p1, p2 = (*f)(&f2), (*f)(&f1)
printmapping(m, p1, p2)
}
func printmapping(m map[*f]s, p1, p2 *f) {
fmt.Println("pointer 1:", m[(*f)(p1)])
fmt.Println("pointer 2:", m[(*f)(p2)])
}
英文:
I recently used a map in one of my golang projects, that had function pointers as keys like this:
map[*functiontype] somestructtype
One of my colleagues said this was a bad idea, so now I am unsure of this being feasible. I initially deemed it ok, because method pointers can be checked for equality and are immutable. Can someone provide some reasoning on that matter?
Complete example:
package main
import "fmt"
type s struct {
string
}
type f func() string
func func1() string { return "func 1" }
func func2() string { return "func 2" }
func main() {
// make two functions and two pointers to them
f1, f2 := func1, func2
p1, p2 := (*f)(&f1), (*f)(&f2)
// make a map of their function pointers
m := make(map[*f]s)
m[p1] = s{"struct 1"}
m[p2] = s{"struct 2"}
// print out the mapping
printmapping(m, p1, p2)
// reverse the pointers and have that printed
p1, p2 = (*f)(&f2), (*f)(&f1)
printmapping(m, p1, p2)
}
func printmapping(m map[*f]s, p1, p2 *f) {
fmt.Println("pointer 1:", m[(*f)(p1)])
fmt.Println("pointer 2:", m[(*f)(p2)])
}
答案1
得分: 6
如果键的类型是指向函数类型的指针(例如*func()
),那么完全没问题,语义与预期一致:相等的指针是相等的键。
然而,查找映射中的值可能不会按照你的预期工作:示例。在这里,&f
获取了局部变量的地址,对于Add
和Find
的不同调用,这个地址是不同的。当然,下面的示例可能会有用:http://play.golang.org/p/F9jyUxzJhz
如果键的类型不是指针,那么这是一个坏主意,因为从Go 1开始就不可能实现。根据语言规范(在线演示):
对于键类型的操作数,比较运算符
==
和!=
必须完全定义;因此,键类型不能是函数、映射或切片。
由于比较函数的相等性是不可判定的问题,因此函数的==
和!=
操作符是未定义的。
英文:
If the key type is a pointer to a function type (such as *func()
) then it's totally fine and the semantics are as expected: equal pointers are equal keys.
However, looking up the values in the map may not work out as you may expect: example. Here &f
takes the address of the local variable, which is never the same for different invocations of Add
and Find
. The following of course could be useful: http://play.golang.org/p/F9jyUxzJhz
If it is not a pointer, it's a bad idea because it's impossible from Go 1 onwards. According to the language specification (live demo):
> The comparison operators == and != must be fully defined for operands of the key type; thus the key type must not be a function, map, or slice.
==
and !=
aren't defined for functions since the problem of comparing functions for equality is undecidable.
答案2
得分: 4
你的问题太抽象了,没有实际意义。给我们一个真实例子的代码。如何创建一个最小、完整和可验证的示例。。
例如,你是指像这样的代码吗?
package main
import "fmt"
type f func(int) int
type s struct{ i int }
func f1(i int) int { return i }
func f2(i int) int { return i * i }
func main() {
p1, p2 := f1, f2
fmt.Println(p1, &p1, p2, &p2)
m := make(map[*f]s)
m[(*f)(&p1)] = s{f1(42)}
m[(*f)(&p2)] = s{f2(42)}
fmt.Println(m)
p1, p2 = f2, f1
fmt.Println(m)
fmt.Println(p1, &p1, p2, &p2)
}
输出:
0x20000 0x1040a120 0x20020 0x1040a128
map[0x1040a120:{42} 0x1040a128:{1764}]
map[0x1040a128:{1764} 0x1040a120:{42}]
0x20020 0x1040a120 0x20000 0x1040a128
英文:
Your question is too abstract to be meaningful. Give us code for a real example. How to create a Minimal, Complete, and Verifiable example..
For example, do you mean something like this?
package main
import "fmt"
type f func(int) int
type s struct{ i int }
func f1(i int) int { return i }
func f2(i int) int { return i * i }
func main() {
p1, p2 := f1, f2
fmt.Println(p1, &p1, p2, &p2)
m := make(map[*f]s)
m[(*f)(&p1)] = s{f1(42)}
m[(*f)(&p2)] = s{f2(42)}
fmt.Println(m)
p1, p2 = f2, f1
fmt.Println(m)
fmt.Println(p1, &p1, p2, &p2)
}
Output:
0x20000 0x1040a120 0x20020 0x1040a128
map[0x1040a120:{42} 0x1040a128:{1764}]
map[0x1040a128:{1764} 0x1040a120:{42}]
0x20020 0x1040a120 0x20000 0x1040a128
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