To specify a JRE by path
Introduction
Generated exe can use a JRE directory specified by a path. The path can be an absolute path or a relative path to 'jvm.dll' on Windows, 'libjvm.so' on Linux in the JRE directory.
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On Windows, 'jvmdll' should be followed by a path to 'jvm.dll', for example:
jvmdll C:\jdk1.5\jre\bin\client\jvm.dll
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On Linux, 'jvmdll' should be followed by a path to 'libjvm.so', for example:
jvmdll /usr/java/jdk/jre/lib/i386/client/libjvm.so
For Copied JRE Directory
If a JRE directory is copied, not installed, jvmdll is useful to let generated exe to use the copied JRE directory.
To Select the Proper Library
In a JRE directory, there may be more than one 'jvm.dll' files ('libjvm.so' on Linux), for example 'client\jvm.dll' and 'server\jvm.dll', jvmdll configure item can be used to specify one of them explicitly.
JRE Find Order
Configure item 'jre_order' is to specify JRE find order. Letter 'c' stands for the JRE specified by jvmdll item.
The default value of jre_order is 'cbsie', so generated exe will use the JRE specified by jvmdll if jvmdll exists. If letter 'c' is not in the value of jre_order, the jvmdll is ignored.
- See jre_order page for details of specifying JRE find order.
Comments
using the jvm.dll in Jar2Exe for a bundled JRE
When creating an exe file from an executable jar and using the bundled jre option, along with selecting the jvm.dll within the "bundled jre folder location" I get a "Java Runtime Environment not valid" error. Obviously, I need the jvm.dll to execute the created exe file using the minimal jre folder contents. How do I do that? All my different entry variations don't work. Really need support to actually provide a step by step guide.
Please check the 32 or 64 bits of JRE
The 32 or 64 bits of JRE need to match 32 or 64 bits of exe file, at the last step of Jar2Exe, https://www.jar2exe.com/usage/wizard/stepoutput, there is an option "Create 64 bits"
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