Newbie question: Why does this work?

[Philipp.Weissenbacher@gmail.com]

Hi all!
This is most certainly a total newbie question, but why doesn't the
following code cause a segfault?

void insertion_sort( int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;
break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}

IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

Thanks in advance and excuse me for asking such a dumb question

[Richard Heathfield]

Re: Newbie question: Why does this work?

Philipp.Weissen bacher@gmail.co m said:

Hi all!
This is most certainly a total newbie question, but why doesn't the
following code cause a segfault?
>
void insertion_sort( int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {

This isn't broken because, first time through, i = 0, so j becomes i - 1 =
-1, which is not >= 0, so the loop body is skipped completely.

cout << a[j] << " <= " << v << endl;

Just for the record, this is a C newsgroup, not a C++ newsgroup.

<snip>

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[santosh]

Re: Newbie question: Why does this work?

Philipp.Weissen bacher@gmail.co m wrote:

Hi all!
This is most certainly a total newbie question, but why doesn't the
following code cause a segfault?
>
void insertion_sort( int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;
break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}
>
IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.
>
Thanks in advance and excuse me for asking such a dumb question

The second for loop will be entered only when it's condition is true,
i.e., only when j is >= 0. This will not happen after the first
iteration of the outer loop. During the second iteration of the outer
loop, j will be initialised to 0 so the code under the inner for loop
will be executed.

[Walter Roberson]

Re: Newbie question: Why does this work?

In article <642e48fe-b105-4178-bac2-c5a49c8e466c@a2 3g2000hsc.googl egroups.com>,
Philipp.Weissen bacher@gmail.co m <Philipp.Weisse nbacher@gmail.c omwrote:

>This is most certainly a total newbie question, but why doesn't the
>following code cause a segfault?

>void insertion_sort( int a[], int length)
>{
int i;
for (i=0; i < length; i++)
{
> int j, v = a[i];
> for (j = i - 1; j >= 0; j--)
> {
> if (a[j] <= v) {
> cout << a[j] << " <= " << v << endl;

The above line will not compile, as you have not defined a variable
named 'cout', nor a variable named 'endl'. Furthermore, left-shifting
a pointer by a value, or a value by a pointer, is not a defined
operation in C. Perhaps in copying out the program for the posting,
you accidently pasted in a line from another window in which you
had something dealing with some other programming language such as
Rogaine ?

> break;
> }
> a[j + 1] = a[j];
> }
> a[j + 1] = v;
}
>}

>IMHO the segfault should occur at line 9 as on the first pass through
>the first for-loop i = o ergo j is -1 causing a[j] to provoke a
>segfault.

When i = 0, and you start the for (j = i - 1; j >= 0; j--) loop,
then j -will- be initialized to -1, but the loop condition j >= 0
will be evaluated before any trips are taken through the loop, and will
be found to be false, so the loop will not be executed. The j >= 0
protects the loop from executing when j is not at least 0.
--
"Man's life is but a jest,
A dream, a shadow, bubble, air, a vapor at the best."
-- George Walter Thornbury

[Default User]

Re: Newbie question: Why does this work?

Philipp.Weissen bacher@gmail.co m wrote:

IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

Even if you had accessed outside the boundaries, that's undefined
behavior. There is no defined behavior for undefined behavior. That
includes segfaults or any other behavior.





Brian

[Charlton Wilbur]

Re: Newbie question: Why does this work?

>>>>"P" == Philipp Weissenbacher@g mail com

>>>><Philipp.We issenbacher@gma il.comwrites:

PHi all! This is most certainly a total newbie question, but why
Pdoesn't the following code cause a segfault?

Because segfaults are a bonus, not a requirement.

Charlton


--
Charlton Wilbur
cwilbur@chromat ico.net

[Philipp.Weissenbacher@gmail.com]

Re: Newbie question: Why does this work?

On 16 Mai, 21:01, Charlton Wilbur <cwil...@chroma tico.netwrote:

>>>"P" == Philipp Weissenbacher@g mail com
>>><Philipp.Wei ssenbac...@gmai l.comwrites:

>
PHi all! This is most certainly a total newbie question, but why
Pdoesn't the following code cause a segfault?
>
Because segfaults are a bonus, not a requirement.
>
Charlton
>
--
Charlton Wilbur
cwil...@chromat ico.net

At first thanks for all the answers. And yes I'm actually using C++; I
just missed some couts. Sorry for that one.
@Roberson: I always thought a for loop does at lest one iteration
before checking the condition. Gotta reread that part ...

[Walter Roberson]

Re: Newbie question: Why does this work?

In article <a235703d-62b9-4378-9518-8b15fc8505a5@d7 7g2000hsb.googl egroups.com>,
Philipp.Weissen bacher@gmail.co m <Philipp.Weisse nbacher@gmail.c omwrote:

>@Roberson: I always thought a for loop does at lest one iteration
>before checking the condition. Gotta reread that part ...

No, the only iteration form that does at least one iteration before
checking the condition is "do until"

--
"I think Walter's legacy will be that of a man with a God-given
ability that got the most out of it at every possible chance."
-- Eddie Payton

[Keith Thompson]

Re: Newbie question: Why does this work?

"Philipp.Weisse nbacher@gmail.c om" <Philipp.Weisse nbacher@gmail.c omwrites:

This is most certainly a total newbie question, but why doesn't the
following code cause a segfault?
>
void insertion_sort( int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;

printf("%d <= %d\n", a[j], v); /* This is comp.lang.c. */

break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}
>
IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

Here's something you could have tried that probably would have avoided
the need to post the question:

By inserting a printf call at the beginning of the for loop:

printf("j = %d\n", j);

you could have seen that j never has the value -1 (which would leave
you to fix the logic problem that prevents the loop from being
executed at all).

Or you could do the equivalent in a debugger; the details of how to do
that are off-topic here, but should be in your debugger's documentation.

In any case, even if you did evaluate a[j] with j == -1, there's no
guarantee that it would cause a seg fault, or any other particular
result. The behavior is simply undefined (i.e., the C standard says
absolutely nothing about what might happen). It could cause a seg
fault, it could quietly access some piece of memory outside the array,
it could clobber something that causes your program to crash later.
It can't *really* make demons fly out of your nose (as the old joke
here goes), but if it did, it wouldn't be a violation of the C
standard.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

Re: Newbie question: Why does this work?

roberson@ibd.nr c-cnrc.gc.ca (Walter Roberson) writes:

In article <a235703d-62b9-4378-9518-8b15fc8505a5@d7 7g2000hsb.googl egroups.com>,
Philipp.Weissen bacher@gmail.co m <Philipp.Weisse nbacher@gmail.c omwrote:
>

>>@Roberson: I always thought a for loop does at lest one iteration
>>before checking the condition. Gotta reread that part ...

>
No, the only iteration form that does at least one iteration before
checking the condition is "do until"

Or, if you happen to be programming in C <OT>or C++</OT>, "do while".
<OT>Pascal has "repeat until"; perhaps that's what you were thinking
of.</OT>

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Walter Roberson]

Re: Newbie question: Why does this work?

In article <lnd4nmghpa.fsf @nuthaus.mib.or g>,
Keith Thompson <kst-u@mib.orgwrote:

>roberson@ibd.n rc-cnrc.gc.ca (Walter Roberson) writes:

>In article <a235703d-62b9-4378-9518-8b15fc8505a5@d7 7g2000hsb.googl egroups.com>,

>No, the only iteration form that does at least one iteration before
>checking the condition is "do until"

>Or, if you happen to be programming in C <OT>or C++</OT>, "do while".
><OT>Pascal has "repeat until"; perhaps that's what you were thinking
>of.</OT>

Nah, I just have a thick head today :(
--
'Roberson' is my family name; my given name is 'Walter'.

[nembo kid]

Re: Newbie question: Why does this work?

Philipp.Weissen bacher@gmail.co m ha scritto:

cout <<

Bleah... ;-)

[Philipp.Weissenbacher@gmail.com]

Re: Newbie question: Why does this work?

On 16 Mai, 21:52, rober...@ibd.nr c-cnrc.gc.ca (Walter Roberson) wrote:

In article <lnd4nmghpa.... @nuthaus.mib.or g>,
Keith Thompson <ks...@mib.orgw rote:
>

rober...@ibd.nr c-cnrc.gc.ca (Walter Roberson) writes:

In article <a235703d-62b9-4378-9518-8b15fc850...@d7 7g2000hsb.googl egroups.com>,
No, the only iteration form that does at least one iteration before
checking the condition is "do until"

Or, if you happen to be programming in C <OT>or C++</OT>, "do while".
<OT>Pascal has "repeat until"; perhaps that's what you were thinking
of.</OT>

>
Nah, I just have a thick head today :(
--
'Roberson' is my family name; my given name is 'Walter'.

@Walter: Thought this would be more polite. I'm a German native-
speaker and we have something called T-V distinction. The English
counterpart to "sie" appears to be calling someone with his family
name (at least for me) whereas calling someone by his given name
indicates a much closer relationship.