what does __leave do?

[Greg Turner]

Could someone run the following code under a microsoft compiler for me, 
and let me know the result?

main() {
  int x = 0;
  __try {
    x = 1;
    __leave;
  } __finally {
    x = 2;
  }
  printf("x was %d\n", x);
}

thanks,

[Francois Gouget]

On Mon, 13 Jan 2003, Greg Turner wrote:
> Could someone run the following code under a microsoft compiler for me,
> and let me know the result?
>
> main() {
>   int x = 0;
>   __try {
>     x = 1;
>     __leave;

x = 3;
> } __finally {
>     x = 2;
>   }
>   printf("x was %d\n", x);
> }

I modified it as shown above and it prints:
x was 2

So '__leave' is like a 'break' but for exception blocks.

[Greg Turner]

On Monday 13 January 2003 01:07 pm, Francois Gouget wrote:
> On Mon, 13 Jan 2003, Greg Turner wrote:
> > Could someone run the following code under a microsoft compiler for
> > me, and let me know the result?
> >
> > main() {
> >   int x = 0;
> >   __try {
> >     x = 1;
> >     __leave;
>   x = 3;
>
> > } __finally {
> >     x = 2;
> >   }
> >   printf("x was %d\n", x);
> > }
>
> I modified it as shown above and it prints:
> x was 2
>
> So '__leave' is like a 'break' but for exception blocks.

cool, quick turnaround on that one.  thanks.