jsii-diff
Version:
Assembly comparison for jsii
171 lines • 6.79 kB
JavaScript
;
Object.defineProperty(exports, "__esModule", { value: true });
exports.isSuperType = isSuperType;
exports.isNominalSuperType = isNominalSuperType;
exports.isStructuralSuperType = isStructuralSuperType;
exports.prependReason = prependReason;
const util_1 = require("./util");
/**
* Check whether A is a supertype of B
*
* Put differently: whether any value of type B would be assignable to a
* variable of type A.
*
* We always check the relationship in the NEW (latest, updated) typesystem.
*/
function isSuperType(a, b, updatedSystem) {
if (a.void || b.void) {
throw new Error('isSuperType() does not handle voids');
}
if (a.isAny) {
return { success: true };
}
if (a.primitive !== undefined) {
if (a.primitive === b.primitive) {
return { success: true };
}
return failure(`${b.toString()} is not assignable to ${a.toString()}`);
}
if (a.arrayOfType !== undefined) {
// Arrays are covariant
if (b.arrayOfType === undefined) {
return failure(`${b.toString()} is not an array type`);
}
return prependReason(isSuperType(a.arrayOfType, b.arrayOfType, updatedSystem), `${b.toString()} is not assignable to ${a.toString()}`);
}
if (a.mapOfType !== undefined) {
// Maps are covariant (are they?)
if (b.mapOfType === undefined) {
return failure(`${b.toString()} is not a map type`);
}
return prependReason(isSuperType(a.mapOfType, b.mapOfType, updatedSystem), `${b.toString()} is not assignable to ${a.toString()}`);
}
// Every element of B can be assigned to A
if (b.unionOfTypes !== undefined) {
const analyses = b.unionOfTypes.map((bbb) => isSuperType(a, bbb, updatedSystem));
if (analyses.every((x) => x.success)) {
return { success: true };
}
return failure(`some of ${b.toString()} are not assignable to ${a.toString()}`, ...(0, util_1.flatMap)(analyses, (x) => (x.success ? [] : x.reasons)));
}
// There should be an element of A which can accept all of B
if (a.unionOfTypes !== undefined) {
const analyses = a.unionOfTypes.map((aaa) => isSuperType(aaa, b, updatedSystem));
if (analyses.some((x) => x.success)) {
return { success: true };
}
return failure(`none of ${b.toString()} are assignable to ${a.toString()}`, ...(0, util_1.flatMap)(analyses, (x) => (x.success ? [] : x.reasons)));
}
// We have two named types, recursion might happen so protect against it.
try {
// For named types, we'll always do a nominal typing relationship.
// That is, if in the updated typesystem someone were to use the type name
// from the old assembly, do they have a typing relationship that's accepted
// by a nominal type system. (That check also rules out enums)
const nominalCheck = isNominalSuperType(a, b, updatedSystem);
if (nominalCheck.success === false) {
return nominalCheck;
}
// At this point, the only thing left to do is recurse into the structs.
// We used to do that here, but we don't anymore; structs check themselves
// for structural weakening/strengthening.
return { success: true };
}
catch (e) {
return failure(e.message);
}
}
/**
* Find types A and B in the updated type system, and check whether they have a supertype relationship in the type system
*/
function isNominalSuperType(a, b, updatedSystem) {
if (a.fqn === undefined) {
throw new Error(`I was expecting a named type, got '${a.toString()}'`);
}
// Named type vs a non-named type
if (b.fqn === undefined) {
return failure(`${b.toString()} is not assignable to ${a.toString()}`);
}
// Short-circuit of the types are the same name, saves us some lookup
if (a.fqn === b.fqn) {
return { success: true };
}
// We now need to do subtype analysis on the
// Find A in B's typesystem, and see if B is a subtype of A'
const B = updatedSystem.tryFindFqn(b.fqn);
const A = updatedSystem.tryFindFqn(a.fqn);
if (!B) {
return failure(`could not find type ${b.toString()}`);
}
if (!A) {
return failure(`could not find type ${a.toString()}`);
}
// If they're enums, they should have been exactly the same (tested above)
// enums are never subtypes of any other type.
if (A.isEnumType()) {
return failure(`${a.toString()} is an enum different from ${b.toString()}`);
}
if (B.isEnumType()) {
return failure(`${b.toString()} is an enum different from ${a.toString()}`);
}
if (B.extends(A)) {
return { success: true };
}
return failure(`${b.toString()} does not extend ${a.toString()}`);
}
/**
* Is struct A a structural supertype of struct B?
*
* Trying to answer the question, is this assignment legal for all values
* of `expr in B`.
*
* ```ts
* const var: A = expr as B;
* ```
*
* A is a structural supertype of B if all required members of A are also
* required in B, and of a compatible type.
*
* Nullable members of A must either not exist in B, or be of a compatible
* type.
*/
function isStructuralSuperType(a, b, updatedSystem) {
// We know all members can only be properties, so that makes it easier.
const bProps = b.getProperties(true);
// Use timing words in error message to make it more understandable
const formerly = b.system === updatedSystem ? 'formerly' : 'newly';
const is = b.system === updatedSystem ? 'is' : 'used to be';
const removed = b.system === updatedSystem ? 'removed' : 'added';
for (const [name, aProp] of Object.entries(a.getProperties(true))) {
const bProp = bProps[name];
if (aProp.optional) {
// Optional field, only requirement is that IF it exists, the type must match.
if (!bProp) {
continue;
}
}
else {
if (!bProp) {
return failure(`${formerly} required property '${name}' ${removed}`);
}
if (bProp.optional) {
return failure(`${formerly} required property '${name}' ${is} optional`);
}
}
const ana = isSuperType(aProp.type, bProp.type, updatedSystem);
if (!ana.success) {
return failure(`property ${name}`, ...ana.reasons);
}
}
return { success: true };
}
function failure(...reasons) {
return { success: false, reasons };
}
function prependReason(analysis, message) {
if (analysis.success) {
return analysis;
}
return failure(message, ...analysis.reasons);
}
//# sourceMappingURL=type-analysis.js.map