I believe there can be an integer overflow, without a silent
wrap-around, in the following example:
int a = INT_MAX;
a++;
Am I right? Could this lead to an abnormal program termination in some
implementations ?
If so, could this happen without an arithmetical operation, i.e. because
of an assignment between different-ranged integers? I mean:
int a = 12345;
char b = a;
Or is it just an arithmetical operation issue?
9  5372 
Mike Wahler wrote: Enrico 'Trippo' Porreca <tr****@lombard iacom.it> wrote in message
news:3F******** ****@lombardiac om.it...
I believe there can be an integer overflow,
Yes, there can. When it happens, the language deems
it 'undefined behavior', which means anything is
allowed to happen.
Perfect. If so, could this happen without an arithmetical operation, i.e. because
of an assignment between different-ranged integers? I mean:
int a = 12345;
char b = a;
If 12345 is outside the range of type 'char', then
data will be lost.
Suppose 12345 is indeed outside the range of 'char' on my platform. Do I
get undefined behaviour then? I guess b can contain almost anything
(from the point of view of the C standard).
Enrico 'Trippo' Porreca wrote:
Mike Wahler wrote: Enrico 'Trippo' Porreca <tr****@lombard iacom.it> wrote in message
news:3F******** ****@lombardiac om.it...
I believe there can be an integer overflow,
Yes, there can. When it happens, the language deems
it 'undefined behavior', which means anything is
allowed to happen.
Perfect.
If so, could this happen without an arithmetical operation, i.e. because
of an assignment between different-ranged integers? I mean:
int a = 12345;
char b = a;
If 12345 is outside the range of type 'char', then
data will be lost.
Suppose 12345 is indeed outside the range of 'char' on my platform. Do I
get undefined behaviour then? I guess b can contain almost anything
(from the point of view of the C standard).
From a "legalistic " standpoint, the Standard permits one
of two outcomes: Either `b' receives an implementation-defined
value, or an implementation-defined signal is raised.
From
a practical standpoint, nearly every implementation takes the
former course, silently delivering some kind of value to `b'.
(It's been thirty years since I last used a machine where the
option of raising a signal might have made sense -- and that
machine didn't have a C implementation. )
Note that the situation is a bit different for unsigned
integral types. `unsigned char c = a;' would not raise any
signal, and would always deliver a value to `c'. The exact
value would be implementation-defined because different
implementations can have differing numbers of bits in their
`char' objects, but once the implementation-specific value
of `UCHAR_MAX' is known, the value stored in `c' can be
predicted with certainty.
-- Er*********@sun .com
Eric Sosman wrote: From a "legalistic " standpoint, the Standard permits one
of two outcomes: Either `b' receives an implementation-defined
value, or an implementation-defined signal is raised.
From
a practical standpoint, nearly every implementation takes the
former course, silently delivering some kind of value to `b'.
(It's been thirty years since I last used a machine where the
option of raising a signal might have made sense -- and that
machine didn't have a C implementation. )
Now it's perfectly clear. Thank you very much.
Enrico 'Trippo' Porreca wrote:
I believe there can be an integer overflow, without a silent
wrap-around, in the following example:
int a = INT_MAX;
a++;
Am I right? Could this lead to an abnormal program termination
in some implementations ?
Yes.
If so, could this happen without an arithmetical operation,
i.e. because of an assignment between different-ranged
integers? I mean:
int a = 12345;
char b = a;
Again, yes. While many implementations silently convert such
overflows into nonsensical values, you cannot assume such. The
actual behaviour is undefined.
--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!
CBFalconer <cb********@yah oo.com> wrote in message news:<3F******* ********@yahoo. com>... Enrico 'Trippo' Porreca wrote:
.... int a = 12345;
char b = a;
Again, yes. While many implementations silently convert such
overflows into nonsensical values, you cannot assume such. The
actual behaviour is undefined.
No. If char is unsigned, the behaviour is perfectly well defined.
If char is signed then the behaviour is implementation defined on C90
implementations .
--
Peter
Peter Nilsson wrote: CBFalconer <cb********@yah oo.com> wrote: Enrico 'Trippo' Porreca wrote:
... int a = 12345;
char b = a;
Again, yes. While many implementations silently convert such
overflows into nonsensical values, you cannot assume such. The
actual behaviour is undefined.
No. If char is unsigned, the behaviour is perfectly well defined.
If char is signed then the behaviour is implementation defined on
C90 implementations .
Touche. chars are different.
--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!
CBFalconer <cb********@yah oo.com> wrote: Peter Nilsson wrote: CBFalconer <cb********@yah oo.com> wrote: > Enrico 'Trippo' Porreca wrote:
... > > int a = 12345;
> > char b = a;
>
> Again, yes. While many implementations silently convert such
> overflows into nonsensical values, you cannot assume such. The
> actual behaviour is undefined.
No. If char is unsigned, the behaviour is perfectly well defined.
If char is signed then the behaviour is implementation defined on
C90 implementations .
Touche. chars are different.
It's nothing to do with chars. Integer overflow causes undefined
behaviour, but a conversion of a value that can't fit into the target
type is implementation-defined behaviour in C90. This has been gone
over on this newsgroup many times.
- Kevin.
Kevin Easton wrote:
CBFalconer <cb********@yah oo.com> wrote: Peter Nilsson wrote: CBFalconer <cb********@yah oo.com> wrote:
> Enrico 'Trippo' Porreca wrote:
...
> > int a = 12345;
> > char b = a;
>
> Again, yes. While many implementations silently convert such
> overflows into nonsensical values, you cannot assume such. The
> actual behaviour is undefined.
No. If char is unsigned, the behaviour is perfectly well defined.
If char is signed then the behaviour is implementation defined on
C90 implementations .
Touche. chars are different.
It's nothing to do with chars. Integer overflow causes undefined
behaviour, but a conversion of a value that can't fit into the target
type is implementation-defined behaviour in C90. This has been gone
over on this newsgroup many times.
Did I say anything else? I simply accepted the correction that an
overflow into a char, be it plain, signed, unsigned, is different
from an integer overflow.
--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!
pete wrote:
CBFalconer wrote:
Kevin Easton wrote:
CBFalconer <cb********@yah oo.com> wrote:
> Peter Nilsson wrote:
>> CBFalconer <cb********@yah oo.com> wrote:
>> > Enrico 'Trippo' Porreca wrote:
>> ...
>> > > int a = 12345;
>> > > char b = a;
You used the word "overflow", when the code in question
merely has the assignement of an out of range value
(assuming that 12345 is out of range for char in this case).
The assignement of an out of range value,
is different from overflow and
the behavior is implementation defined, not undefined.
.... and also assuming that char is signed.
If unsigned, then the behavior is defined.
--
pete This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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