I have installed the Chocolatey Java package choco install jre8 -y
which installs both the 32-bit and 64-bit packages.
When I type java.exe
on the console, I see that it is the 32-bit version and that it is located at C:\Program Files (x86)\Common Files\Oracle\Java\javapath\java.exe
. When I go to that PATH location, it seems to be a soft or hard link, and it contains java.exe
, javaw.exe
, and javaws.exe
.
I find all of this quite confusing. Can someone explain some of these things please?
- What is the
javapath
folder is (well, it's not a folder, looks like a link of some kind, but how and why is it setup this way)?
- Why is it defaulting to the 32-bit version of java?
- What if I need the 64-bit version of java, do I have to manually type in the path to the 64-bit version of Java or else it will use the 32-bit version? Or does java someone work out if an application prefers 64-bit java and then it automatically uses that runtime? And does the
javapath
being a non-folder have something to do with this?
- There is also a strange folder at
C:\Program Files (x86)\Common Files\Oracle\Java\javapath_target_2324328
which also contains the java.exe
, javaw.exe
, and javaws.exe
, so it gets even more confusing, what is that for?