Re: Class.forName().newInstance() vs new

From:
Patricia Shanahan <pats@acm.org>
Newsgroups:
comp.lang.java.programmer
Date:
Sat, 11 Jun 2011 11:01:10 -0700
Message-ID:
<6aadnUFKX6n1MG7QnZ2dnUVZ_uKdnZ2d@earthlink.com>
On 6/11/2011 10:13 AM, Abu Yahya wrote:

I have a class that instantiates a private member based on a value in a
configuration file.

The value is used to determine the class name.

I initially decided to use code similar to the following:

// class name comes from property file
String dbMgrClassName = props.getProperty("db.manager");
Class<?> k = Class.forName(dbMgrClassName);
db = (DB) k.newInstance();
db.init(props);

However, a colleague pointed out that using "new" is faster than using
"Class.forName().newInstance()". So, I changed the code to the following:

String dbMgrClassName = props.getProperty("db.manager");
if (dbMgrClassName.equals("DB2"))
db = new DB2();
else if (dbMgrClassName.equals("Oracle"))
db = new Oracle();
else if (dbMgrClassName.equals("SQLServer"))
db = new SQLServer();
else { // handle unrecognized
Class<?> k = Class.forName(dbMgrClassName);
db = (DB) k.newInstance();
}
db.init(props);

My question is: Does using if-else statements like the above really
improve performance? What is the best approach to instantiate classes
whose type you don't know at compile time?


Is performance really a critical issue here? Instantiating a database
manager is usually done quite infrequently, normally once per run, and
is often followed by opening a database, a relatively heavyweight operation.

I could see a case for the second version if you wanted to control the
options for the implementing class. Given the decision to preserve the
old code to handle unrecognized class names it does not even have that
advantage. In order to properly test the second version you will need an
additional class, not "DB2", "Oracle", or "SQLServer", so that the
fall-through code gets exercised.

The first version is simpler and easier to test, so it wins unless it
makes a significant difference in overall program performance.

Patricia

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