java - try-finally block clarification -


this question has answer here:

when try execute following function in java:

public static int myfunc (int x) {     try {         return x;     } {         x++;     }        } public static void main (string args[]) {     int y=5,z;     z = myfunc(y);     system.out.println(z); }

the output printed on console 5, 1 expect 6 printed. idea why?

...would expect 6 printed. idea why?

the x++ happens after value 5 has been read x. remember follows return statement expression. when return statement reached, expression evaluated (its value determined), , resulting value used function's return value. in myfunc, here's order of happens:

  1. enter try block.
  2. evaluate expression x (e.g., value of x).
  3. set value function's return value.
  4. enter finally block.
  5. increment x.
  6. exit function using return value step 3.

so of when leave function, though x 6, return value determined earlier. incrementing x doesn't change that.

your code in myfunc analogous this:

int y = x; x++;

there, read value of x (just return x does), assign y, , increment x. y unaffected increment. same true return value of function.

it might clearer functions: assuming foo function outputs "foo" , returns 5, , bar function outputs "bar", code:

int test() {     try {         return foo();     }     {         bar();     } }

...executes foo, outputting "foo", executes bar, outputting "bar", , exits test return value 5. don't expect function wait call foo until after finally occurs (that strange), , indeed doesn't: calls when reaches return foo(); statement, because evaluates expression foo(). same true of return x;: evaluates expression , remembers result of statement.


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