Because you need that code to execute regardless of any exceptions that may be thrown. For example, you may need to clean up some unmanaged resource (the 'using' construct compiles to a try/finally block).
What happens if an exception you're not handling gets thrown? (I hope you're not catching Throwable...)
What happens if you return from inside the try block?
What happens if the catch block throws an exception?
A finally block makes sure that however you exit that block (modulo a few ways of aborting the whole process explicitly), it will get executed. That's important for deterministic cleanup of resources.
Note that (in Java at least, probably also in C#) it's also possible to have a try block without a catch, but with a finally. When an exception happens in the try block, the code in the finally block is run before the exception is thrown higher up:
InputStream in = new FileInputStream("somefile.xyz");
try {
somethingThatMightThrowAnException();
}
finally {
// cleanup here
in.close();
}
You may want to put the code that you want to anyway get executed irrespective of what happens in your try or catch block.
Also if you are using multiple catch and if you want to put some code which is common for all the catch blocks this would be a place to put- but you cannot be sure that the entire code in try has been executed.
Even though our application is closed forcefully there will be some tasks, which we must execute (like memory release, closing database, release lock, etc), if you write these lines of code in the finally block it will execute whether an exception is thrown or not...
Your application may be a collection of threads, Exception terminates the thread but not the whole application, in this case finally is more useful.
In some cases finally won't execute such as JVM Fail, Thread terminate, etc.
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
console.writeline(ex.Message);
}
finally
{
console.writeline("Finally block");
}
console.writeline("After finally");
what would be printed in the above scenario?
Yes guessed it right:
ex.Message--whatever it is (probably attempted division by zero)
Finally block
After finally
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
throw(ex);
}
finally
{
console.writeline("Finally block");
}
console.writeline("After finally");
What would this print? Nothing! It throws an error since the catch block raised an error.
In a good programming structure, your exceptions would be funneled, in the sense that this code will be handled from another layer. To stimulate such a case i'll nested try this code.
try
{
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
throw(ex);
}
finally
{
console.writeline("Finally block")
}
console.writeline("After finally");
}
catch(Exception ex)
{
console.writeline(ex.Message);
}
In this case the output would be:
Finally block
ex.Message--whatever it is.
It is clear that when you catch an exception and throw it again into other layers(Funneling), the code after throw does not get executed. It acts similar to just how a return inside a function works.
You now know why not to close your resources on codes after the catch block.Place them in finally block.