Hi, the following is compiled in 9.3, glibc-2.3.4-23.4, and gcc-3.3.5-5. The result is: mask=0 mask1=4294967295

I would expect mask = mask1 = 0?

Regards, Verdi

=========================================================== #include

int main() { unsigned int mask = (unsigned int) -1 >> 32;

int i = 0; unsigned int mask1 = ((unsigned int) -1) >> (32 - i);

printf("mask=%u mask1=%u\n", mask, mask1); return 0; } =========================================================== Marcus' comment that this is an undefined behavior is absolutely correct (ISO C90 6.5.7), so if your code relies on mask1 (or mask) resulting in 0 is

On Wednesday 20 July 2005 5:57 am, Verdi March wrote: probably a bad assumption and the code should be corrected. Apparently what is happening in the second case is that ((unsigned int) -1) is being changed to a signed int, and the sign bit is getting propagated. BTW: On a 64-bit system (Itanium) the values are both 0. -- Jerry Feldman Boston Linux and Unix user group http://www.blu.org PGP key id:C5061EA9 PGP Key fingerprint:053C 73EC 3AC1 5C44 3E14 9245 FB00 3ED5 C506 1EA9

On Thursday 21 July 2005 4:46 pm, Philipp Thomas wrote:

Jerry Feldman [Wed, 20 Jul 2005 10:29:44 -0400]:

...

On Wednesday 20 July 2005 5:57 am, Verdi March wrote:

BTW: On a 64-bit system (Itanium) the values are both 0.

Of cause! You'd obviously have to use

unsigned int mask = (unsigned int) -1 >> 64; and unsigned int mask1 = ((unsigned int) -1) >> (64 - i); True since on a 64-bit system (unsigned int) -1 might be promoted to an unsigned long, but I'm too lazy to look at the promotion rules in the standard right now. -- Jerry Feldman Boston Linux and Unix user group http://www.blu.org PGP key id:C5061EA9 PGP Key fingerprint:053C 73EC 3AC1 5C44 3E14 9245 FB00 3ED5 C506 1EA9