15122 Compile-Time Step 2: Determine Method Signature

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EXPRESSIONS

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If F has the form G<, Yk-1, extends U, Yk+1, >, where U involves Tj, then if A has a supertype that is one of:

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G<, Xk-1, V, Xk+1, >, where V is a type expression Then this algorithm is applied recursively to the constraint V << U

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Again, let s keep things as simple as possible, and consider only the case where G has a single type argument A <<F in this case means A << G< extends U> As above, it must be the case that A is a subtype of some invocation of G However, A may now be a subtype of either G<V>, or G< extends V>, or G< super V> We examine these cases in turn The rst variation is described (generalized to multiple arguments) by the sub-bullet directly above We therefore have A = G<V> << G< extends U> The rules of subtyping for wildcards imply that V << U

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Extending the analysis above, we have A = G< extends V> << G< extends U> The rules of subtyping for wildcards again imply that V << U

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Here, we have A = G< super V> << G< extends U> In general, we cannot conclude anything in this case However, it is not necessarily an error It may be that U will eventually be inferred to be Object, in which case the call may indeed be valid Therefore, we simply refrain from placing any constraint on U

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G<, Xk-1, extends V, Xk+1, > Then this algorithm is applied recursively to the constraint V << U

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Otherwise, no constraint is implied on Tj

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EXPRESSIONS

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Compile-Time Step 2: Determine Method Signature 15122

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If F has the form G<, Yk-1, super U, Yk+1, >, where U involves Tj, then if A has a supertype that is one of:

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G<, Xk-1, V, Xk+1, > Then this algorithm is applied recursively to the constraint V >> U

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As usual, we consider only the case where G has a single type argument A <<F in this case means A << G< super U> As above, it must be the case that A is a subtype of some invocation of G A may now be a subtype of either G<V>, or G< extends V>, or G< super V> We examine these cases in turn The rst variation is described (generalized to multiple arguments) by the sub-bullet directly above We therefore have A = G<V> << G< super U> The rules of subtyping for wildcards imply that V >> U

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We have A = G< super V> << G< super U> The rules of subtyping for lowerbounded wildcards again imply that V >> U

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Here, we have A = G< extends V> << G< super U> In general, we cannot conclude anything in this case However, it is not necessarily an error It may be that U will eventually be inferred to the null type, in which case the call may indeed be valid Therefore, we simply refrain from placing any constraint on U

Otherwise, no constraint is implied on Tj

G<, Xk-1, super V, Xk+1, > Then this algorithm is applied recursively to the constraint V >> U

15122 Compile-Time Step 2: Determine Method Signature

EXPRESSIONS

Otherwise, no constraint is implied on Tj

Otherwise, if the constraint has the form A = F

DISCUSSION

If F = Tj, then the constraint Tj = A is implied

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

DISCUSSION

Clearly, if the array types U[] and V[] are the same, their component types must be the same

If F has the form G<, Yk-1, U, Yk+1, >, 1 k n where U is type expression that involves Tj, then if A is of the form G<, Xk-1, V, Xk+1,> where V is a type expression, this algorithm is applied recursively to the constraint V = U If F has the form G<, Yk-1, extends U, Yk+1, >, where U involves Tj, then if A is one of: G<, Xk-1, extends V, Xk+1, > Then this algorithm is applied recursively to the constraint V = U Otherwise, no constraint is implied on Tj

If F has the form G<, Yk-1, super U, Yk+1 ,>, where U involves Tj, then if A is one of:

Such a constraint is never part of the initial constraints However, it can arise as the algorithm recurses We have seen this occur above, when the constraint A << F relates two parameterized types, as in G<V> << G<U>