Re: C++ exception context
On Apr 17, 1:05 am, "jg" <jgu...@gmail.com> wrote:
Can someone explain what the standard says about
exception context ? For example, suppose bar() always throw
an exception, is the value of g undefined or 1 after the handler
(empty in this case) returns ?
foo()
{
int g=0;
try {
g++;
bar(); // bar() throw an exception
} catch (...)
}
printf("g=%d\n", g);
}
In this case, no. The usual C++ rules apply: values will
reflect all changes before the last sequence point, and none of
the changes after the next one. Anything which changes with no
intervening sequence point is up in the air. Throwing an
exception is a sequence point, but don't forget that the order
of evaluation of an expression is not specified, so something
like the following can be very problematic:
f( g++, bar() ) ;
If bar() throws, g may or may not have been incremented.
Be careful, too, because sequence points only define a partial
ordering. For example, given:
f( auto_ptr( new Toto ), bar() ) ;
one possible ordering is:
tmp = new Toto
bar()
auto_ptr( tmp )
If bar() throws, you've got a memory leak.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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