Hi Everyone. Thanks for the help with the qudratic equation problem...I
didn't think about actually doing the math...whoops. Anyway... I'm
having some trouble getting the following program to work. I want to
output a bit pattern from base 10 input. All I get is a zero after the
input...I've looked over the code but can't see the problem...any ideas?
/* display the bit pattern corresponding to
a signed decimal integer */
#include<stdio. h>
#include<stdlib .h>
int main(void)
{
int a, b, m, count, nbits;
unsigned mask;
/* determine the word size in bits and set the initial mask */
nbits = 8 * sizeof(int);
m = 0x1 << (nbits - 1); /* Place 1 in leftmost position */
/* main loop */
do {
/* read a signed integer */
printf("\n\nEnt er an integer value (0 to stop): ", a);
scanf("%d", &a);
/* output the bit pattern */
mask = m;
for(count = 1; count <= nbits; count++); {
b = (a & mask) ? 1 : 0; /* set display bit on or off */
printf("%x", b); /* print display bit */
if (count % 4 == 0)
printf(" "); /* blank space after ever 4th digit */
mask >>= 1; /* shift mask 1 position to the right */
}
} while (a != 0);
return EXIT_SUCCESS; /* If it's there why not use it people! */
}
Nov 14 '05
47  5291 
In article <qs************ @brenda.flash-gordon.me.uk>,
Flash Gordon <sp**@flash-gordon.me.uk> wrote: If you had:
if ((c=b)==f(c))
or:
if ((a=c)==f(c))
you would have undefined behaviour.
The second one has no undefined behavior.
On Thu, 09 Sep 2004 19:40:51 GMT
CBFalconer <cb********@yah oo.com> wrote: Flash Gordon wrote:
... snip ...
If you had:
if ((c=b)==f(c))
Yes
or:
if ((a=c)==f(c))
No.
you would have undefined behaviour.
In the latter case c is being accessed only to extract its value,
not to alter it. Assuming it is not a volatile value.
Sorry, I flipped languages for a moment to one that allows pass by
reference.
--
Flash Gordon
Sometimes I think shooting would be far too good for some people.
Although my email address says spam, it is real and I read it.
On Thu, 9 Sep 2004, Flash Gordon wrote:
so************@ yahoo.com (Mark Piffer) wrote: so************@ yahoo.com (Mark Piffer) wrote... Ok, how about the following:
int a,b,c;
int f(int);
if( (a=b) == f(c) ) { // undefined behaviour?
}
As the order of execution is unspecified (is it?), the evaluation
sequence could be:[1] a=b
[2] f(c) // sequence point (according to 6.5.2.2/10)
[3] == // using the result of [1] and [2]
[...] Can someone enlighten me about this? Either I got ignored or the
posting vanished too soon from the servers, but I think the question
isn't trivial (ok, for regulars here maybe it is) and certainly not
OT.
The order of evaluation of a=b and f(c) does not affect the result in
this case, so why would it be undefined behaviour?
Because the Standard says so. Dan Pop's response is correct, as usual;
and most of the other replies are wrong. If you're having trouble
figuring out why the Standard says it's UB, it might help to consider the
perfectly reasonable function definition
int f(int x) { a = x; }
(Then again, it might not. ;-)
Admittedly the value
of a might not be updated until after the comparison has been done, but
the comparison still has to be done against the correct value.
Not if it's undefined. :)
If you had:
if ((c=b)==f(c))
or:
if ((a=c)==f(c))
you would have undefined behaviour.
Those examples aren't any different from the original example; in fact,
they're just special cases of the original (the first where a:=:c and the
second where b:=:c).
HTH,
-Arthur
"Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote: so************@ yahoo.com (Mark Piffer) wrote...
if( (a=b) == f(c) ) { // undefined behaviour?
As the order of execution is unspecified (is it?), the evaluation
sequence could be:[1] a=b
[2] f(c) // sequence point (according to 6.5.2.2/10)
[3] == // using the result of [1] and [2]
The order of evaluation of a=b and f(c) does not affect the result in
this case, so why would it be undefined behaviour?
Because the Standard says so. Dan Pop's response is correct, as usual;
and most of the other replies are wrong. If you're having trouble
figuring out why the Standard says it's UB, it might help to consider the
perfectly reasonable function definition
int f(int x) { a = x; }
How about this:
int c;
memset(x, c = 0, sizeof x);
In article <84************ **************@ posting.google. com>, ol*****@inspire .net.nz (Old Wolf) writes:
int c;
memset(x, c = 0, sizeof x);
That looks OK to me. N869 6.5.16p4 says that accessing the result of
the assignment operator *after the next sequence point* causes UB;
that's what made
if ((a=b) == f(c))
undefined (because there's a sequence point after the arguments to f
are evaluated). In your example, the result of the assignment
operator (in "c = 0") is accessed before the sequence point that
occurs when all of memset's arguments have been evaluated, because
accessing it is part of evaluating memset's arguments; hence no UB.
So, curiously enough,
if ((a=b) == c)
is legal, but eg
if ((a=b) == (c,d))
is not, because of the sequence point caused by evaluating the first
operand of the comma operator.
OK, now someone should point out where I got it wrong...
(I'm glad Mark re-posted his question. This was a tricky one, and
I certainly hadn't noticed it before.)
--
Michael Wojcik mi************@ microfocus.com
The lark is exclusively a Soviet bird. The lark does not like the
other countries, and lets its harmonious song be heard only over the
fields made fertile by the collective labor of the citizens of the
happy land of the Soviets. -- D. Bleiman
In article <ch*********@ne ws3.newsguy.com >, mw*****@newsguy .com (Michael Wojcik) wrote: In article <84************ **************@ posting.google. com>,
ol*****@inspire .net.nz (Old Wolf) writes:
int c;
memset(x, c = 0, sizeof x);
That looks OK to me. N869 6.5.16p4 says that accessing the result of
the assignment operator *after the next sequence point* causes UB;
that's what made
if ((a=b) == f(c))
undefined (because there's a sequence point after the arguments to f
are evaluated). In your example, the result of the assignment
operator (in "c = 0") is accessed before the sequence point that
occurs when all of memset's arguments have been evaluated, because
accessing it is part of evaluating memset's arguments; hence no UB.
So, curiously enough,
if ((a=b) == c)
is legal, but eg
if ((a=b) == (c,d))
is not, because of the sequence point caused by evaluating the first
operand of the comma operator.
OK, now someone should point out where I got it wrong...
That has been discussed to some length in comp.std.c.
You are not "accessing" the value of the assignment operator. You can
only "access" lvalues. The result of (a=b) is not an lvalue. You can use
it without accessing it.
In article <ch************ *************** ******@slb-newsm1.svr.pol. co.uk>, Christian Bau <ch***********@ cbau.freeserve. co.uk> writes:
You are not "accessing" the value of the assignment operator. You can
only "access" lvalues. The result of (a=b) is not an lvalue. You can use
it without accessing it.
Wouldn't that render N869 6.5.16p4 meaningless? It reads:
If an attempt is made to modify the result of an assignment
operator or to access it after the next sequence point, the
behavior is undefined.
If "access" only applies to lvalues, and the result of an assignment
operator is never an lvalue, then this statement describes a case
that can't occur.
Was this corrected in the final version of C99?
--
Michael Wojcik mi************@ microfocus.com
Thus, the black lie, issuing from his base throat, becomes a boomerang to
his hand, and he is hoist by his own petard, and finds himself a marked man.
-- attributed to a "small-town newspaper editor in Wisconsin"
In article <ci********@new s2.newsguy.com> , mw*****@newsguy .com (Michael Wojcik) wrote: In article <ch************ *************** ******@slb-newsm1.svr.pol. co.uk>,
Christian Bau <ch***********@ cbau.freeserve. co.uk> writes:
You are not "accessing" the value of the assignment operator. You can
only "access" lvalues. The result of (a=b) is not an lvalue. You can use
it without accessing it.
Wouldn't that render N869 6.5.16p4 meaningless? It reads:
If an attempt is made to modify the result of an assignment
operator or to access it after the next sequence point, the
behavior is undefined.
If "access" only applies to lvalues, and the result of an assignment
operator is never an lvalue, then this statement describes a case
that can't occur.
Please read the start of this thread again where I gave an example. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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