«

unsigned x = 0;
if(x-- < 4)
printf("foo\n");
else
printf("bar\n");

GCC for Data General Aviion 5000 (an 88000 machine) will convert this
into:

unsigned x = 0;
if(--x < 3) ....

in the assembly.

This seems broken if x == ~0.

Robert Samelson wrote:
) unsigned x = 0;
) if(x-- < 4)
) printf("foo\n");
) else
) printf("bar\n");
)
) GCC for Data General Aviion 5000 (an 88000 machine) will convert this
) into:
)
) unsigned x = 0;
) if(--x < 3) ....

That doesn't look like assembly to me.
I assume this is your 'decompiling' of the actual assembly instructions.
It means that there either is a fault in the compiler's compilation, or
in your decompilation. Without information on what the actual assembly
instructions are, it's hard to say which it is.

) in the assembly.
)
) This seems broken if x == ~0.

You should test what the actual output from the code is, to see if it
does what it's supposed to.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

On Mon, 13 Feb 2012 18:53:49 +0000 (UTC), Robert Samelson
<> wrote:

>unsigned x = 0;
>if(x-- < 4)
> printf("foo\n");
>else
> printf("bar\n");
>
>GCC for Data General Aviion 5000 (an 88000 machine) will convert this
>into:
>
> unsigned x = 0;
> if(--x < 3) ....
>
>in the assembly.
>
>This seems broken if x == ~0.

As long as the code produces the correct response for any initial
value of x, why would you think it broken? For x initialized to 3, 2,
or 1, it should print foo. For all other others it should print bar.
Do you have results that show otherwise?

--
Remove del for email

On 02/13/2012 03:14 PM, Barry Schwarz wrote:
> On Mon, 13 Feb 2012 18:53:49 +0000 (UTC), Robert Samelson
> <> wrote:
>
>> unsigned x = 0;
>> if(x-- < 4)
>> printf("foo\n");
>> else
>> printf("bar\n");
>>
>> GCC for Data General Aviion 5000 (an 88000 machine) will convert this
>> into:
>>
>> unsigned x = 0;
>> if(--x < 3) ....
>>
>> in the assembly.
>>
>> This seems broken if x == ~0.
>
> As long as the code produces the correct response for any initial
> value of x, why would you think it broken? For x initialized to 3, 2,
> or 1, it should print foo. For all other others it should print bar.

No, for x initialized with 0 (as shown), the original C code should
print "foo\n", not "bar\n". What you've described corresponds to the
second sample of C code. That difference is why the rewrite would be
incorrect.

Robert Samelson <> writes:
> unsigned x = 0;
> if(x-- < 4)
> printf("foo\n");
> else
> printf("bar\n");
>
> GCC for Data General Aviion 5000 (an 88000 machine) will convert this
> into:
>
> unsigned x = 0;
> if(--x < 3) ....
>
> in the assembly.
>
> This seems broken if x == ~0.

It's broken if and only if the program produces incorrect output. C
code specifies behavior, not machine code.

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
Will write code for food.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Feb 13, 6:53*pm, Robert Samelson <r...@invalid.com> wrote:
> unsigned x = 0;
> if(x-- < 4)
> * * * * * * * * printf("foo\n");
> else
> * * * * * * * * printf("bar\n");
>
> GCC for Data General Aviion 5000 (an 88000 machine) will convert this
> into:
>
> * * * * unsigned x = 0;
> * * * * if(--x < 3) ....
>
> in the assembly.
>
> This seems broken if x == ~0.
>
Yes, it will fail, because of wrap round. If it prints "bar" than this
is a compiler bug.

I was convinced once that gcc generated wrong code
because my program ran differently on two machines.

But further inspection revealed that I was using the %
operator on a zero operand.