Google Maps will look even more like Waze



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Google Maps is, for many users, the best application for navigation and GPS location. However, Waze also occupies a place of choice among so many others. In addition, both applications belong to the Google giant.

This is the fifth major addition to the Google Maps service since the beginning of this year and was previously proposed in mid-2018. It will be, for all practical purposes, another function of

 Google, Google Maps, Waze, Google Maps, navigation

Meanwhile, we have already learned that Google Maps would identify areas usually controlled by radars. These two features were unique to Waze, but they will now be present in Waze.

Another feature of Waze that will reach Google Maps

both. However, we will still have another function to access Google Maps.

According to the new report of the BGR portal, we will also have the opportunity to report a road accident. This is one of the biggest strengths of the Waze app and, according to the initial report of a Reddit user, would also arrive at Maps.

In this social network, the user u / oldkingcoles noticed this feature tested in the service. Google Maps. Something that, for all practical purposes, is a mirror of what we have already found in the aforementioned Waze

 Google, Maps, Waze, Google Maps, Navigation

Therefore, in addition to be able to share the location of a speed camera, we can also report the occurrence of a traffic accident. All of this is clear, baduming this report is reliable.

If that happens, we will have a stronger social component, not just Waze.

An irreversible convergence between navigation services?

Yet despite the fact that the new function has been seen, it does not mean that it is definitely implemented. For now, it is in test phase and can reach or not the stable channel, accessible to all users.

Finally, Waze has been owned by Google for over 5 years, but not at all. it's no longer curious to see the competition between the two apps. (F, b, e, v, n, t, s) {if (f.fbq) returns; n = f.fbq = function () {n.callMethod?
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