langer@gibbs.uchicago.edu (Steve Langer) (04/23/91)
Hi --
What is the expected output from the following program? Is it defined?
#include <stdio.h>
main() {
printf("1 << 32 = %o\n", ((unsigned) 1) << 32);
}
Running on a Sun 4 gives 1.
Running on a Sun 3 gives 0.
Running on an Iris gives 1.
K&R (first edition) page 45 says that left shifts fill vacated bits by
0, which does not seem to be happening.
If the shift is done one step at a time the answer is 0 on the Sun 4 and
the Iris, as expected.
#include <stdio.h>
main() {
int i;
unsigned x = 1;
for(i=0; i<32; i++) x <<= 1;
printf("%o\n", x);
}
E-mail responses are welcome!
Thanks in advance,
Steve
langer@control.uchicago.eduhenry@zoo.toronto.edu (Henry Spencer) (04/23/91)
In article <1991Apr22.225641.1122@midway.uchicago.edu> langer@control.uchicago.edu (Steve Langer) writes: >What is the expected output from the following program? Is it defined? > printf("1 << 32 = %o\n", ((unsigned) 1) << 32); Assuming you're on a 32-bit machine, no, it is undefined. Translation, expect different machines to do different things, some of them unpleasant. -- And the bean-counter replied, | Henry Spencer @ U of Toronto Zoology "beans are more important". | henry@zoo.toronto.edu utzoo!henry
steve@taumet.com (Stephen Clamage) (04/24/91)
langer@gibbs.uchicago.edu (Steve Langer) writes: >What is the expected output from the following program? Is it defined? >#include <stdio.h> >main() { > printf("1 << 32 = %o\n", ((unsigned) 1) << 32); >} It is not defined, so the variation in outputs you got is legal. In 3.3.7, the Standard says "If the value of the right operand [of the shift operator] ... is greater than or equal to the width in bits of the promoted left operand, the behavior is undefined." -- Steve Clamage, TauMetric Corp, steve@taumet.com
enag@ifi.uio.no (Erik Naggum) (04/24/91)
In article <1991Apr23.003314.5194@zoo.toronto.edu>, Henry Spencer writes: In article <1991Apr22.225641.1122@midway.uchicago.edu>, Steve Langer writes: >What is the expected output from the following program? Is it defined? > printf("1 << 32 = %o\n", ((unsigned) 1) << 32); Assuming you're on a 32-bit machine, no, it is undefined. Translation, expect different machines to do different things, some of them unpleasant. In concrete terms: (1) Some machines treat the shift as "modulo word-length", i.e. 1 << 32 => 1 << 0 => 1. (2) Some machines treat the shift as an unsigned integer, and the result as "modulo 2**wordlength", i.e. 1 << 32 => 4294967296 => 0. (3) Some machines do something else when the shift operand is larger than the applicable word-size, such as laugh at you. -- [Erik Naggum] <enag@ifi.uio.no> Naggum Software, Oslo, Norway <erik@naggum.uu.no>
jfc@athena.mit.edu (John F Carr) (04/25/91)
In article <ENAG.91Apr24181029@maud.ifi.uio.no> enag@ifi.uio.no (Erik Naggum) writes: >(3) Some machines do something else when the shift operand is larger > than the applicable word-size, such as laugh at you. Which machines and compilers make (1 << wordsize) neither 1 nor 0? -- John Carr (jfc@athena.mit.edu)
enag@ifi.uio.no (Erik Naggum) (04/27/91)
In article <ENAG.91Apr24181029@maud.ifi.uio.no>, Erik Naggum writes: (3) Some machines do something else when the shift operand is larger than the applicable word-size, such as laugh at you. In article <1991Apr25.153729.2360@athena.mit.edu>, John F Carr writes: Which machines and compilers make (1 << wordsize) neither 1 nor 0? Alternative three was added for the sake of completeness. I don't know of any, but it's conceivable that this is an overflow condition. -- [Erik Naggum] <enag@ifi.uio.no> Naggum Software, Oslo, Norway <erik@naggum.uu.no>