Function compareTypeParametersIdentical

test/fixtures/snapshot/typescript.js:58159–58181

(sourceParams, targetParams)

            return result || ts.emptyArray;
        }
        function compareTypeParametersIdentical(sourceParams, targetParams) {
            if (ts.length(sourceParams) !== ts.length(targetParams)) {
                return false;
            }
            if (!sourceParams || !targetParams) {
                return true;
            }
            var mapper = createTypeMapper(targetParams, sourceParams);
            for (var i = 0; i < sourceParams.length; i++) {
                var source = sourceParams[i];
                var target = targetParams[i];
                if (source === target)
                    continue;
                // We instantiate the target type parameter constraints into the source types so we can recognize `<T, U extends T>` as the same as `<A, B extends A>`
                if (!isTypeIdenticalTo(getConstraintFromTypeParameter(source) || unknownType, instantiateType(getConstraintFromTypeParameter(target) || unknownType, mapper)))
                    return false;
                // We don't compare defaults - we just use the type parameter defaults from the first signature that seems to match.
                // It might make sense to combine these defaults in the future, but doing so intelligently requires knowing
                // if the parameter is used covariantly or contravariantly (so we intersect if it's used like a parameter or union if used like a return type)
                // and, since it's just an inference _default_, just picking one arbitrarily works OK.
            }
            return true;
        }
        function combineUnionThisParam(left, right, mapper) {
            if (!left || !right) {
                return left || right;