Cannot use type assertion on type parameter value
In PHP programming, type assertion is a common technique used to determine the type of a variable. However, sometimes we may encounter a problem: type assertions cannot be used on type parameter values. This question may confuse some developers. In this article, we will explore the causes of this problem and provide some solutions. Whether we are beginners or experienced developers, by understanding and solving this problem, we can better understand and use type assertions. Let’s take a look!
Question content
We cannot use type assertions for generic type variables. This does seem strange considering that it is a common behavior allowed by interface{}
but not constrained by interface{}
. Wondering if there are any workarounds?
// This works func isInt(x interface{}) bool { _, ok := x.(int) return ok; } // Compile Error // invalid operation: cannot use type assertion on type parameter // value x (variable of type T constrained by interface{}) func isInt2[T interface{}](x T) bool { _, ok := x.(int) return ok; }
Solution
tl;Doctor
You can only perform type assertions on interface values. Therefore, you must first convert x
to a valid interface type, in this case any
/ interface{}
:
func isint[t any](x t) (ok bool) { _, ok = any(x).(int) // convert, then assert return }
So why does this compile fail?
_, ok = x.(int) // ... cannot use type assertion on type parameter value ...
x
t
is a type parameter, not an interface. It is only bounded by the interface. The go (revision 1.18
) language specification clearly states that type parameters are not allowed to appear in type assertions:
For expressions of interface type but not type x
Parameters , and type t
...The symbol x.(t)
is called a type assertion.
Also from Generics Tutorial, learn why parameter types need to be resolved at compile time:
Although constraints on type parameters usually represent a set of Type, at compile time the type parameter represents a single type – The type supplied by the calling code as a type parameter. if type The type of the parameter is not allowed by the type parameter constraint, The code cannot be compiled.
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