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Compile-Time Step 2: Determine Method Signature 15122
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G<, Xk-1, super V, Xk+1, > Then this algorithm is applied recursively to the constraint V = U Otherwise, no constraint is implied on Tj
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Otherwise, no constraint is implied on Tj
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Otherwise, if the constraint has the form A >> F
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Such situations arise when the algorithm recurses, due to the contravariant subtyping rules associated with lower-bounded wildcards (those of the form G< super X>) It might be tempting to consider A>> F as being the same as F << A, but the problem of inference is not symmetric We need to remember which participant in the relation includes a type to be inferred
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We do not make use of such constraints in the main body of the inference algorithm However, they are used in section 151228
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This follows from the covariant subtype relation among array types The constraint A >> F, in this case means that A >> U[] A is therefore necessarily an array type V[], or a type variable whose upper bound is an array type V[] - otherwise the relation A >> U[] could never
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If F = Tj, then the constraint Tj <: A is implied
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If F = U[], where the type U involves Tj, then if A is an array type V[], or a type variable with an upper bound that is an array type V[], where V is a reference type, this algorithm is applied recursively to the constraint V >> U Otherwise, no constraint is implied on Tj
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15122 Compile-Time Step 2: Determine Method Signature
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hold true It follows that V[] >> U[] Since array subtyping is covariant, it must be the case that V >> U
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If F has the form G<, Yk-1, U, Yk+1, >, where U is a type expression that involves Tj, then:
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In this case (once again restricting the analysis to the unary case), we have the constraint A >> F = G<U> A must be a supertype of the generic type G However, since A is not a parameterized type, it cannot depend upon the type argument U in any way It is a supertype of G<X> for every X that is a valid type argument to G No meaningful constraint on U can be derived from A
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Our goal here is to simplify the relationship between A and F We aim to recursively invoke the algorithm on a simpler case, where the actual type argument is known to be an invocation of the same generic type declaration as the formal Let s consider the case where both H and G have only a single type argument Since we have the constraint A = H<X> >> F = G<U>, where H is distinct from G, it must be the case that H is some proper superclass or superinterface of G There must be a (non-wildcard) invocation of H that is a supertype of F = G<U> Call this invocation V If we replace F by V in the constraint, we will have accomplished the goal of relating two invocations of the same generic (as it happens, H) How do we compute V The declaration of G must introduce a formal type parameter S, and there must be some (non-wildcard) invocation of H, H<U1>, that is a supertype of
If A is an invocation of a generic type declaration H, where H is either G or superclass or superinterface of G, then: If H G , then let S1, , Sn be the formal type parameters of G, and let H<U1, , Ul> be the unique invocation of H that is a supertype of G<S1 , , Sn>, and let V = H<U1, , Ul>[Sk = U] Then, if V :> F this algorithm is applied recursively to the constraint A >> V
If A is an instance of a non-generic type, then no constraint is implied on Tj
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Compile-Time Step 2: Determine Method Signature 15122
Otherwise, if A is of the form G<, Xk-1, W, Xk+1, >, where W is a type expression this algorithm is applied recursively to the constraint W = U