Create an instance of a generic type using a pointer to a type parameter

php Xiaobian Yuzai will introduce in this article how to use pointers to type parameters to create instances of generic types. Generics are a concept widely used in programming that allow us to write general code without specifying specific types. In PHP, although there is no native support for generics, we can simulate generics by using pointers to type parameters. This method can make us more flexible and efficient when writing code, and improve the reusability and maintainability of code. In the following content, we will detail how to use this method to create instances of generic types and give some examples of practical applications.

Question content

Given these type definitions:

type n interface{ ~int | ~float32 | ~float64 }

type s[t any] struct {
    t t
}

type myint int

type pspmyint[t myint] *s[*t]
type spmyint[t *myint,] s[t]
type spmyint2[t myint] s[*t]

I can create a var of type pspmyint

func createps[t myint]() pspmyint[t] {
    var i t
    s := s[*t]{t: &i}
    return &s
}

But I don't know how to create a variable of spmyint or spmyint2.

this

func createSP[T myInt]() spMyInt2[T] {
    var i T
    s := S[*T]{t: &i}
    return s
}

Compilation failed You cannot use s (variable of type s[*t]) as type spmyint2[t] in the return statement.

Solution

First, do not use exact type parameter constraints. This makes almost no sense. When you declare a function as createps[t myint](), the type parameter type set has cardinality 1, so it can effectively be instantiated by myintonly and are always . You can rewrite the function like this:

65bee32705632

solve this problem:

Type s[*t] is different from spmyint2[t]. However, since the underlying type of spmyint2[t] is s[*t], you can simply convert: 65bee32705638

As for

type spmyint[t *myint,] s[t] (where the comma is not a typo but is needed to avoid parsing ambiguity), things are not that simple.

The problem is that the type parameter

is not its type constraint. Therefore, a type literal cannot be used to instantiate a different unnamed type literal. What needs to be made clear is:

// naive attempt that doesn't compile
func createsp1[t myint]() spmyint[*t] {
    var i t
    s := s[*t]{t: &i}
    return spmyint[*t](s)
}

You might think that

spmyint[t *myint] has a type parameter constrained to *myint and that the function t is constrained to the base type myint Constraints, Therefore *t should satisfy phpcnc phpcnt * myint. This is incorrect, for apparently non-obvious reasons, the type literal *t is not equal to *myint. So actually you can't write a generic constructor for type spmyint[t *myint,] s[t].

But you are lucky, because the cardinality of type constraints is one. So you can remove the type parameter:

func createSP1() spMyInt[*myInt] {
    var i myInt
    s := S[*myInt]{t: &i}
    return spMyInt[*myInt](s)
}

The above is the detailed content of Create an instance of a generic type using a pointer to a type parameter. For more information, please follow other related articles on the PHP Chinese website!