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  5292 
William L. Bahn wrote: "fb" <fb@goaway.ne t> wrote in message
news:dRvZc.2857 14$M95.236771@p d7tw1no...
Derrick Coetzee wrote:
fb wrote:
nbits = 8 * sizeof(int);
The error is on this line.
right-o thanks.
So....that told you what the problem is?
I thought it might have...I was going to check it later, though at first
glance in did seem fine to me (assuming 8 bit bytes).
Watch out - you're hand waving it. I get the impression you
aren't trying to understand what the problem is, you are just
wanting someone to tell you what code to throw at the compiler.
umm...I don't want to admit that...but it's kinda true. I had been
staring at it for a while and gave up...posted this message and left.
Otherwise you should have asked what could possibly be wrong with
that line. It should work for you - but it is not robust and,
conceivably, might not work on your machine - and even if it did
have a problem it wouldn't yield the result you are experiencing.
But I'm not getting the impression that understanding the
relationship between your code and the results your code produce
is very high on your priority list.
I can't say that I cared at that point in time, though a good nights
sleep fixed that...I'll take a few breaths and think about it next time.
ciao.
On Thu, 02 Sep 2004 01:06:38 -0400, Derrick Coetzee wrote: Mac wrote:
mask >>= 1; /* shift mask 1 position to the right */
I don't think the above line will do what you expect.
Nice try invoking UD on signed shift, but mask is unsigned. It was
merely assigned from a signed variable, but I think that's well-defined
with two's-complement semantics. As an aside, in my experience, even
signed variable shifts typically get compiled into logical rather than
arithmetic shifts.
You are right that I was assuming mask was an int. I didn't mention UB,
though. I just quoted from the standard, and the part I quoted said
implementation defined. But since mask is unsigned, the part I quoted
(which you snipped) is irrelevant.
--Mac
In <ch************ *************@t heriver.com> Barry Schwarz <sc******@deloz .net> writes: On Thu, 02 Sep 2004 09:41:24 GMT, pete <pf*****@mindsp ring.com> wrote:
William L. Bahn wrote:
As assignments between signed and unsigned data types of the same
basic type required to copy the bit pattern if the value in the
expression can't be represented in the type of the object being
assigned to? Or is this undefined behavior?
Assigning an out of range value to a signed integer,
is implementation defined.
No, it is undefined.
Chapter and verse, please.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
On 6 Sep 2004 16:13:24 GMT, Da*****@cern.ch (Dan Pop) wrote: In <ch************ *************@t heriver.com> Barry Schwarz <sc******@deloz .net> writes:
On Thu, 02 Sep 2004 09:41:24 GMT, pete <pf*****@mindsp ring.com> wrote:
William L. Bahn wrote:
As assignments between signed and unsigned data types of the same
basic type required to copy the bit pattern if the value in the
expression can't be represented in the type of the object being
assigned to? Or is this undefined behavior?
Assigning an out of range value to a signed integer,
is implementation defined.
No, it is undefined.
Chapter and verse, please.
I was going from n869 J.2 "Conversion to or from an integer type
produces a value outside the range that can be represented (6.3.1.4)."
but I see that applies only to floating to integer conversions and not
integer to integer. 6.3.1.3 covers integer to integer and specifies
implementation defined.
<<Remove the del for email>>
Could someone please explain what section 6.5.16-4 means
"The order of evaluation of the operands is unspecified. 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."
I understand the order of evaluation part.
d = b*b-4*a*c;
if (d>0) ...
This is pretty common code and obviously not undefined behavior if the
variables are properly defined and assigned values. But it seems that
the if statement attempts to access the result of the previous
assignment operator after the end of statement sequence point as
described in the quote. What am I misinterpreting ?
<<Remove the del for email>>
On Mon, 6 Sep 2004, Barry Schwarz wrote:
Could someone please explain what section 6.5.16-4 means
"The order of evaluation of the operands is unspecified. 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."
I understand the order of evaluation part.
d = b*b-4*a*c;
if (d>0) ...
This is pretty common code and obviously not undefined behavior if the
variables are properly defined and assigned values. But it seems that
the if statement attempts to access the result of the previous
assignment operator after the end of statement sequence point as
described in the quote. What am I misinterpreting ?
The result of the assignment operator is discarded. It never gets
used. What gets used in the 'if' statement is the value of the object
'd'. (That 'd' was assigned to in the previous statement has absolutely
no relevance.)
The condition
if ((d = b*b-4*a*c) > 0) ...
uses the result of the assignment operator in an expression, but it
uses it /before/ the next sequence point, so it's okay.
HTH,
-Arthur
In article <ch************ *************@t heriver.com>,
Barry Schwarz <sc******@deloz .net> wrote: Could someone please explain what section 6.5.16-4 means
"The order of evaluation of the operands is unspecified. 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."
I understand the order of evaluation part.
d = b*b-4*a*c;
if (d>0) ...
This is pretty common code and obviously not undefined behavior if the
variables are properly defined and assigned values. But it seems that
the if statement attempts to access the result of the previous
assignment operator after the end of statement sequence point as
described in the quote. What am I misinterpreting ?
It is not accessing the "result of the assignment operator", it is
accessing the variable d. You have to do some pretty weird things to
"access the result of an assignment operator":
typedef struct { int x [3]; } mystruct;
mystruct a, b;
int* p;
p = &(a = b).x[2];
*p = 3;
Normally, if you have
int i = 3, j = 4;
i = j;
the result of the assignment operator is not "j", and it is not "i", but
the number 4. In the example above, the "result of the assignment
operator" is a value of type mystruct, but it is not a or b. I managed
to get the address of an element of that struct into p; p is not equal
to &a.x[2] or &b.x[2]. The assignment *p = 3; is undefined behavior.
"Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote in message news:<Pi******* *************** ************@un ix49.andrew.cmu .edu>... On Mon, 6 Sep 2004, Barry Schwarz wrote:
Could someone please explain what section 6.5.16-4 means
"The order of evaluation of the operands is unspecified. 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."
I understand the order of evaluation part.
d = b*b-4*a*c;
if (d>0) ...
This is pretty common code and obviously not undefined behavior if the
variables are properly defined and assigned values. But it seems that
the if statement attempts to access the result of the previous
assignment operator after the end of statement sequence point as
described in the quote. What am I misinterpreting ?
The result of the assignment operator is discarded. It never gets
used. What gets used in the 'if' statement is the value of the object
'd'. (That 'd' was assigned to in the previous statement has absolutely
no relevance.)
Can you please elaborate why the result of assignment operator is
discarded ? i even test the above condition by writing a program,
but the behaviour was as expected. The value assigned to 'd' was
not discarded. So, when and where such scenarios would cause undefined
behaviour.
thanx in advance for any help/hints....
The condition
if ((d = b*b-4*a*c) > 0) ...
uses the result of the assignment operator in an expression, but it
uses it /before/ the next sequence point, so it's okay.
HTH,
-Arthur
Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote in message news:<ch******* *************** ***********@slb-newsm1.svr.pol. co.uk>... In article <ch************ *************@t heriver.com>,
Barry Schwarz <sc******@deloz .net> wrote:
Could someone please explain what section 6.5.16-4 means
"The order of evaluation of the operands is unspecified. 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."
You have to do some pretty weird things to
"access the result of an assignment operator":
typedef struct { int x [3]; } mystruct;
mystruct a, b;
int* p;
p = &(a = b).x[2];
*p = 3;
Normally, if you have
int i = 3, j = 4;
i = j;
the result of the assignment operator is not "j", and it is not "i", but
the number 4. In the example above, the "result of the assignment
operator" is a value of type mystruct, but it is not a or b. I managed
to get the address of an element of that struct into p; p is not equal
to &a.x[2] or &b.x[2]. The assignment *p = 3; is undefined behavior.
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]
Does some other rule catch this construct?
Mark ju**********@ya hoo.co.in (junky_fellow) wrote: "Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote in message news:<Pi******* *************** ************@un ix49.andrew.cmu .edu>... On Mon, 6 Sep 2004, Barry Schwarz wrote:
d = b*b-4*a*c;
if (d>0) ...
The result of the assignment operator is discarded. It never gets
used. What gets used in the 'if' statement is the value of the object
'd'. (That 'd' was assigned to in the previous statement has absolutely
no relevance.)
Can you please elaborate why the result of assignment operator is
discarded ?
Because there's nowhere for it to go to. The statement has finished.
i even test the above condition by writing a program,
but the behaviour was as expected. The value assigned to 'd' was
not discarded.
No, but that's not the same value. It _has_ the same value, but it's not
the same value. Perhaps another example is in order. Say,
x = y = 5;
You'll agree that this is one statement, consisting of two assignment
operations, right? Ok, first of all, = binds to the right, so x = y = 5
is x = (y = 5), not (x = y) = 5. With that in mind, what happens is:
- 5 is evaluated. Its value is, no surprise there, 5.
- the right assignment operation assigns the value of its right operand,
5, to its left operand, y, which now also has the value of 5.
- The right assignment operation _also_ passes the value of its right
operand on, up the evaluation tree.
- The left assignment operation assigns the value of its right
operand, which just happens to be the value of the right assignment,
which in turn is the value of _its_ right operand, which was 5 (ok,
enough whiches already!), to its left operand, x. Now x also has
a value of 5.
Note that by this time the value of y does not matter any more. If y
is volatile, and gets changed behind the program's back between the
two assignments, that does not matter a jot. What gets assigned to x
is _not_ the value of y, but the value of the right assignment.
- The right assignment also passes the value of its right operand up the
evaluation tree.
- Since there is no further operator to use this value, however, it is
discarded. The statement is now complete, and there is a sequence
point here.
So, when and where such scenarios would cause undefined behaviour.
In very convoluted situations, since you need to access a temporary
value in a previous statement, which has already finished. I'm not even
sure Christian's example is valid ISO C; even if it is, it's not _good_
ISO C.
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