Is this broken or wot?

[Robert Samelson]

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.

 

[Willem]

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

 

[Barry Schwarz]

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?

 

[James Kuyper]

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.

 

[Keith Thompson]

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.

 

[Malcolm McLean]

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.

 

[Bl0ckeduser .]

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.