Searched refs:largest (Results 1 – 8 of 8) sorted by relevance
72 long largest = 0; in onDataLoaded() local82 largest = Math.max(size, largest); in onDataLoaded()87 sortAndAddPreferences(items, largest); in onDataLoaded()182 private void sortAndAddPreferences(List<AppItem> items, long largest) { in sortAndAddPreferences() argument185 int percentTotal = largest != 0 ? (int) (items.get(i).total * 100 / largest) : 0; in sortAndAddPreferences()
367 long largest = 0; in bindStats() local386 largest = accumulate(managedKey, knownItems, bucket, in bindStats()387 AppItem.CATEGORY_USER, items, largest); in bindStats()412 largest = accumulate(collapseKey, knownItems, bucket, category, items, largest); in bindStats()436 final int percentTotal = largest != 0 ? (int) (items.get(i).total * 100 / largest) : 0; in bindStats()488 Bucket bucket, int itemCategory, ArrayList<AppItem> items, long largest) { in accumulate() argument499 return Math.max(largest, item.total); in accumulate()
92 public Size largest; field in ResolutionUtil.ResolutionBucket186 Size largest = sizes.get(0); in pickUpToThree() local187 result.add(largest); in pickUpToThree()188 Size lastSize = largest; in pickUpToThree()190 double targetArea = Math.pow(.5, result.size()) * area(largest); in pickUpToThree()
73 `DnsTlsQueryMap` assigns each new query the ID number one greater than the largest75 and usually small. If the largest possible ID number is already in use,
3107 * Returns the index of the largest element along an axis.5638 * Finds values and indices of the k largest entries for the last dimension.5661 * largest elements along each last dimensional slice.
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