Remove sub-namespaces from all remaining typedefs

This commit is contained in:
Peter Robins
2016-06-09 09:21:23 +00:00
parent 09202cf95d
commit 2c29512c80
47 changed files with 492 additions and 479 deletions

View File

@@ -75,14 +75,14 @@ ol.source.ImageVector = function(options) {
/**
* User provided style.
* @type {ol.style.Style|Array.<ol.style.Style>|ol.style.StyleFunction}
* @type {ol.style.Style|Array.<ol.style.Style>|ol.StyleFunction}
* @private
*/
this.style_ = null;
/**
* Style function for use within the library.
* @type {ol.style.StyleFunction|undefined}
* @type {ol.StyleFunction|undefined}
* @private
*/
this.styleFunction_ = undefined;
@@ -188,7 +188,7 @@ ol.source.ImageVector.prototype.getSource = function() {
/**
* Get the style for features. This returns whatever was passed to the `style`
* option at construction or to the `setStyle` method.
* @return {ol.style.Style|Array.<ol.style.Style>|ol.style.StyleFunction}
* @return {ol.style.Style|Array.<ol.style.Style>|ol.StyleFunction}
* Layer style.
* @api stable
*/
@@ -199,7 +199,7 @@ ol.source.ImageVector.prototype.getStyle = function() {
/**
* Get the style function.
* @return {ol.style.StyleFunction|undefined} Layer style function.
* @return {ol.StyleFunction|undefined} Layer style function.
* @api stable
*/
ol.source.ImageVector.prototype.getStyleFunction = function() {
@@ -284,7 +284,7 @@ ol.source.ImageVector.prototype.renderFeature_ = function(feature, resolution, p
* it is `null` the layer has no style (a `null` style), so only features
* that have their own styles will be rendered in the layer. See
* {@link ol.style} for information on the default style.
* @param {ol.style.Style|Array.<ol.style.Style>|ol.style.StyleFunction|undefined}
* @param {ol.style.Style|Array.<ol.style.Style>|ol.StyleFunction|undefined}
* style Layer style.
* @api stable
*/