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This question gave me tough time back a while.
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.