unsigned wierdness

[tobiasoed@hotmail.com]

Hello!
I have the following piece of code:

#include <stdio.h>

int main(void){

unsigned int u;
int i1, i2;
double d1,d2;

u = 0;

i1 = u - 1;
i2 = (signed int) u - 1;

d1 = u - 1;
d2 = (signed int) u - 1;

printf("i1 = %d, i2 = %d, d1 = %f, d2 = %f\n",i1,i2,d1, d2);
}

and don't understand the output:
tobsbox:~/C$ cc -std=c99 -Wall -W -pedantic unsigned.c && ./a.out
i1 = -1, i2 = -1, d1 = 4294967295.0000 00, d2 = -1.000000

It looks like u is promoted to signed int for i2 but not d2. Is
this normal or is it a compiler bug?
Tobias
ps: My compiler is
cc (GCC) 3.4.3 20050227 (Red Hat 3.4.3-22.fc3)
Copyright (C) 2004 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is
NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR
PURPOSE.

[tobiasoed@hotmail.com]

Re: unsigned wierdness


tobias...@hotma il.com wrote:[color=blue]
>
> It looks like u is promoted to signed int for i2 but not d2.[/color]

Ooups, I meant i1 and d1 of course
Tobias

[tobiasoed@hotmail.com]

Re: unsigned wierdness


tobias...@hotma il.com wrote:[color=blue]
> Hello!
> I have the following piece of code:
>
> #include <stdio.h>
>
> int main(void){
>
> unsigned int u;
> int i1, i2;
> double d1,d2;
>
> u = 0;
>
> i1 = u - 1;
> i2 = (signed int) u - 1;
>
> d1 = u - 1;
> d2 = (signed int) u - 1;
>
> printf("i1 = %d, i2 = %d, d1 = %f, d2 = %f\n",i1,i2,d1, d2);
> }
>
> and don't understand the output:
> tobsbox:~/C$ cc -std=c99 -Wall -W -pedantic unsigned.c && ./a.out
> i1 = -1, i2 = -1, d1 = 4294967295.0000 00, d2 = -1.000000[/color]

Never mind. The whole rhs is done in unsigned arithmetic and gives
-1U which is 4294967295 here. The assignement to i1 'overflows' and
happens to give -1. The assignement to d1 doesn't and preserves the
value. I mistakenly thought that the rhs would all be done in
signed.
Tobias.

[Old Wolf]

Re: unsigned wierdness

tobias...@hotma il.com wrote:[color=blue]
> tobias...@hotma il.com wrote:[color=green]
> >
> > int main(void){
> >
> > unsigned int u;
> > int i1, i2;
> > double d1,d2;
> >
> > u = 0;
> > i1 = u - 1;
> > i2 = (signed int) u - 1;
> > d1 = u - 1;
> > d2 = (signed int) u - 1;
> > printf("i1 = %d, i2 = %d, d1 = %f, d2 = %f\n",i1,i2,d1, d2);
> > }
> >
> > i1 = -1, i2 = -1, d1 = 4294967295.0000 00, d2 = -1.000000[/color]
>
> Never mind. The whole rhs is done in unsigned arithmetic and
> gives -1U which is 4294967295 here. The assignement to i1
> 'overflows' and happens to give -1.[/color]

To be precise: since 4294967295 is outside the range of
signed int, the assignment to i1 either gives an
implementation-defined result, or raises an implementation-
defined signal.