Question

I recently downloaded Tomcat 7.x as a zip. Running the version.bat gives the following:

c:\apache-tomcat-7.0.19\bin>version
Using CATALINA_BASE:   "C:\apache-tomcat-7.0.19"
Using CATALINA_HOME:   "c:\apache-tomcat-7.0.19"
Using CATALINA_TMPDIR: "C:\apache-tomcat-7.0.19\temp"
Using JRE_HOME:        "C:\Program Files (x86)\Java\jdk1.6.0_29"
Using CLASSPATH:       "c:\apache-tomcat-7.0.19\bin\bootstrap.jar;C:\apache-tomcat-    7.0.19\bin\tomcat-juli.jar"
Server version: Apache Tomcat/7.0.19
Server built:   Jul 13 2011 11:32:28
Server number:  7.0.19.0
OS Name:        Windows Server 2008 R2
OS Version:     6.1
Architecture:   x86
JVM Version:    1.6.0_29-b11
JVM Vendor:     Sun Microsystems Inc.

Since it’s using the 32 bit version of JRE, is it a safe assumption the Tomcat itself is 32-bit?

Answer 1

The Windows distributions contain executables and a DLL to run Tomcat as a service. You can unzip the download & run Dependency Walker (free) or dumpbin.exe (comes with MS Visual Studio) on the executable to see which processor architecture they support.

See this question for more details: In windows,how do we identify whether a file is 64 bit or 32 bit?

Java programs aren’t 32-bit or 64-bit as native programs are. They run in a virtual machine with a standard architecture. Only the JRE, which implements the virtual machine, is 32-bit or 64-bit.