TITLE: switch statements and variable initialization


PROBLEM: emzee@telerama.lm.com

switch(x) {
   case 1:
     int y = x;	[ modified from original ]
     ...
     break;
   case 2:

etc...

gave me the message: initilization of y missed by case stmt or 
some such thing.

However,

switch(x) {
   case 1:
     {
     int y=x;		[ modified from original ]
     ...
     break;
     }
   case 2:

compiles and executes fine!



RESPONSE: 76001.2500@compuserv.com (Scott J. Baierl)

It doesn't have anything to do with brackets.  


RESPONSE: clamage@Eng.Sun.COM (Steve Clamage), 13 Oct 95

Yes, it does.


RESPONSE: SJB

?All execution paths of a 
switch statement must have the same variable declarations, so they must be 
declared before the switch.

RESPONSE: SC

No, there is no such rule.

The C++ rule is that you cannot jump past the initialization of a variable
unless you bypass the entire block containing the initialization. (See the
ARM, p86 and p91.) The original example did not have that problem, and
should not have caused a compiler complaint, so I modified it slightly to
demonstrate the rule. (I assume the original article did not copy
the offending code exactly, since it was also missing the needed left
brace after the "switch" headers.)

In the first example, taking the 'case 2' leg would leave variable 'y'
in scope, but without having been initialized. That would violate the rule,
so the code is invalid.

In the second example, taking the 'case 2' leg bypasses the entire block
containing 'y', so no rule is violated.