Text 41210, 104 rader
Skriven 2006-10-13 16:11:00 av Michiel van der Vlist (2:280/5555.1)
  Kommentar till text 41157 av mark lewis (1:3634/12.0)
Ärende: smart compilers (was: pi)
=================================
Hello mark.

12 Oct 06 08:56, you wrote to me:

 MvdV>>>> if ((2+2)==5) printf("pi equals 3.12415");
[..]
 MvdV>>> Right! It will not print anythingh at all! If it is a smart
 MvdV>>> compiler - and some are - it won't even make it into the
 MvdV>>> object code.

 ml>> that's interesting... wonder what it (the smart compiler) would
 ml>> do if the "short-circut boolean" flag was tripped the other way?

 MvdV>> That is a meaningless question. ((2+2)==5) will never evaluate
 MvdV>> to TRUE. Unless the compiler is broken and then anything can
 MvdV>> happen.

 ml> no, it is not a meaningless question... you're talking code, now...

No, *I* was just considering this specific example.

 ml> however, for your explicit example, yeah, i'll give it to ya...

Ok, now that we got that out of the way...

 ml> what would your compiler do with each of the following?

 ml>   if (2+2==5) or (3+3==7) then print "egads!";

I have done quit a bit in Pascal programming, but that was a long time ago.
Before the  IBM PC (clones) dominated the market. My compiler went to the
museum together with my home brew 8 bit system running 6809 FLEX. Haven't
written a line of Pascal in over fifteen years. So maybe someone will point out
that it will not complie at all because something is missing...


 ml>   if (2+2==5) and (3+3==7) then print "EGADS!";

 ml>   if (2+2==4) or (3+3==7) then print "whew!";

 ml>   if (2+2==5) or (3+3==6) then print "Whew!";

 ml>   if (2+2==4) and (3+3==6) then print "WHEW!";

All these examples are variations of one ande the same thing. A constant
between the if and the then. The constant is either TRUE or it is FALSE. If it
is true the print statement will be compiled and executed when the object code
is run. If it is FALSEe, the smart compiler will skip it so nothing will ever
be printed.

You want me to do the leg work? Ok...

EGADS! : skipped
whew!  : printed
Whew!  : printed
WHEW!  : printed


 ml> now, try these...

 ml>   a=2; b=2; c=5; d=3; e=3; f=7;
 ml>   if (a+b==c) or (d+e==f) then print "egads!";

Ah, that is a bit more tough. If it is a *very* smart optimising compiler it
will be aware that although the terms in the if statement are written as
variables, the actual values are already known at compile time and so the
compiler will handle it the same as when the actual values were substituted:

if (2+2==5) or (3+3==7) then print "egads!"

The compiler will skip it and nothing will be printed

If it is a less smart compiler, it will treat the terms in the if statement as
true variables - like the ones that are only entered at run time - and generate
the complete object code needed to handle the general case.

Still, when the object code is run, it will evaluate, the condition as FALSE
(2+2 is not equal to 5 and 3+3 inot equal to 7) and it will not look what is
after the then and just skip to the statement following the print "egads!"

 ml>   if (a+b==c) and (d+e==f) then print "EGADS!";

In this example, when the object code is run, is will not evaluate the (d+e==f)
part as the condition has already failed on the first part.  Whatever is after
the and can never make the condition come TRUE, so no need to waste time
evaluating it.

 ml>   c=4;
 ml>   if (a+b==c) or (d+e==f) then print "whew!";

 ml>   c=5; f=6;
 ml>   if (a+b==c) or (d+e==f) then print "Whew!";

 ml>   c=4; f=6;
 ml>   if (a+b==c) and (d+e==f) then print "WHEW!";

I will leave it to the reader to work out the rest, I think I have explained
the basics.


Cheers, Michiel

--- GoldED/W32 3.0.1
 * Origin: Michiel's laptop, on the move somewhere.... (2:280/5555.1)