loop scope

[zahy[dot]bnaya[At]gmail[dot]com]

Hello all,
Sorry for the stupid question, It's probably has a simple solution but
I'm stuck on this for quite a while...

I have this piece of code, I've numbered it for clarification.
As far as I can see it's a nested loop, the first one is (1 to 5 ) and
the inner one is (3 to 4) BUT when I debug it it looks like the inner
loop
is (3-5). why? there are no curly braces so I assumed the 3rd line for
scope is only line 4 and the if scope is line 4.
now, althogh the 5th line "seems" to execute in the debugger the
"hasval" stays 0 no matter what is the value of "digit" , can it be it
is not
really executes?
1. for (i = 0; i < PatternSize-1; i++) /* for each remaining
position in permutation */
2. {digit = perm[i] ; /* initially digit is
value in permutation */
3. for (j = 0; j < i; j++) /* for each previous element in
permutation */
4. if (perm[j] < perm[i]) digit--; /* previous position
contains smaller value, decrement digit*/
5. hashval = hashval * (SIZE - i-1) + digit;} /* multiply digit by
appropriate factor */

Thanks.

[werasm]

> 1. for (i = 0; i < PatternSize-1; i++) /* for each remaining

position in permutation */
Just a quick guess: Should condition be <PatternSize-1> or just
PatternSize, as you're already using less_than "<". The last item would
be missed?

2. {digit = perm[i] ; /* initially digit is
value in permutation */
3. for (j = 0; j < i; j++) /* for each previous element in
permutation */
4. if (perm[j] < perm[i]) digit--; /* previous position
contains smaller value, decrement digit*/
5. hashval = hashval * (SIZE - i-1) + digit;} /* multiply digit by
appropriate factor */

Thanks.

[Ben Pope]

zahy[dot]bnaya[At]gmail[dot]com wrote:

Hello all,
Sorry for the stupid question, It's probably has a simple solution but
I'm stuck on this for quite a while...

I have this piece of code, I've numbered it for clarification.
As far as I can see it's a nested loop, the first one is (1 to 5 ) and
the inner one is (3 to 4) BUT when I debug it it looks like the inner
loop
is (3-5). why? there are no curly braces so I assumed the 3rd line for
scope is only line 4 and the if scope is line 4.
now, althogh the 5th line "seems" to execute in the debugger the
"hasval" stays 0 no matter what is the value of "digit" , can it be it
is not
really executes?
1. for (i = 0; i < PatternSize-1; i++) /* for each remaining
position in permutation */
2. {digit = perm[i] ; /* initially digit is
value in permutation */
3. for (j = 0; j < i; j++) /* for each previous element in
permutation */
4. if (perm[j] < perm[i]) digit--; /* previous position
contains smaller value, decrement digit*/
5. hashval = hashval * (SIZE - i-1) + digit;} /* multiply digit by
appropriate factor */

That looks horrible when posted to the newsgroup. Your comments add
very little to the code. You should add a general comment above the
first for loop that explains what it does.

It's also incomplete code.

If I add braces and indent it properly I get:

for (i = 0; i < PatternSize-1; ++i) {
digit = perm[i];
for (j = 0; j < i; j++) {
if (perm[j] < perm[i]) {
digit--;
}
hashval = hashval * (SIZE - i-1) + digit;
}
}

It obvious to see what you've done.

Indent properly and use braces. It's good for your health.

Ben Pope
--
I'm not just a number. To many, I'm known as a string...

[Ben Pope]

Ben Pope wrote:

zahy[dot]bnaya[At]gmail[dot]com wrote:

1. for (i = 0; i < PatternSize-1; i++) /* for each remaining
position in permutation */
2. {digit = perm[i] ; /* initially digit is
value in permutation */
3. for (j = 0; j < i; j++) /* for each previous
element in
permutation */
4. if (perm[j] < perm[i]) digit--; /* previous position
contains smaller value, decrement digit*/
5. hashval = hashval * (SIZE - i-1) + digit;} /* multiply digit by
appropriate factor */

If I add braces and indent it properly I get:

for (i = 0; i < PatternSize-1; ++i) {
digit = perm[i];
for (j = 0; j < i; j++) {
if (perm[j] < perm[i]) {
digit--;
}
hashval = hashval * (SIZE - i-1) + digit;
}
}

It obvious to see what you've done.

Actually I don't think that is equivalent to what you wrote... Just
stick the braces on, it's so much easier.

If you want to post code here, please make it complete, and minimal, so
that we can paste it into our compilers.

Ben Pope
--
I'm not just a number. To many, I'm known as a string...

[werasm]

That looks horrible when posted to the newsgroup. Your comments add
very little to the code. You should add a general comment above the
first for loop that explains what it does.

I have to agree, I think that is why he was "stuck on this for quite a
while..." :-)

[zahy[dot]bnaya[At]gmail[dot]com]

well....
Thank you all for your "compliments" but It's not my code!
I just have to use it, I agree, It's horrible! (much worse then you
think)
Can anyone define the scopes of the loops relative to the line numbers
(In it's current, horrible state)?

Thanks.

[werasm]

zahy[dot]bnaya[At]gmail[dot]com wrote:

well....
Thank you all for your "compliments" but It's not my code!
I just have to use it, I agree, It's horrible! (much worse then you
think)
Can anyone define the scopes of the loops relative to the line numbers
(In it's current, horrible state)?

Thanks.

Replace the code by this:

for (i = 0; i < PatternSize-1; i++)
{ //Scope of for1 begin
digit = perm[i] ;
for (j = 0; j < i; j++)
{//Scope of for2 begin
if (perm[j] < perm[i])
{//begin
digit--;
}//end
}//Scope of for2 end
hashval = hashval * (SIZE - i-1) + digit; //xxx!
} //Scope of for1 end

1) <PatternSize-1> in the outer for(for1) looks fishy!
2) Make sure line marked <xxx!> is doing what you
want it to do - Note Precedence! I suppose its right.

W

[zahy[dot]bnaya[At]gmail[dot]com]

Again, I did not write this code and the code is doing what it should.
I just want to revise it.
I was expecting the behavior exactly as described by werasm,
But when running it from the debugger it looks like the line:
hashval = hashval * (SIZE - i-1) + digit;
is being executed as part of for2.
However the hashval variable remains at it's intial value (0) although
digit is changing . It looks like the line is not being
executed at all.
How can this be? can the debugger "lie"?
Hope my question is clearer now..
Thanks.

[Hans]

zahy[dot]bnaya[At]gmail[dot]com wrote:

Again, I did not write this code and the code is doing what it should.
I just want to revise it.
I was expecting the behavior exactly as described by werasm,
But when running it from the debugger it looks like the line:
hashval = hashval * (SIZE - i-1) + digit;
is being executed as part of for2.
However the hashval variable remains at it's intial value (0) although
digit is changing . It looks like the line is not being
executed at all.
How can this be? can the debugger "lie"?
Hope my question is clearer now..
Thanks.

Are you compiling with optimization enabled? If so, the debugger will
typically have great difficulty displaying the correct "current source
line", due to how the code has been reshuffled.

If your debugger, supports it, maybe you should take a peek at the
generated assembly code; that will tell you which code is executed in
which loops.

[werasm]

zahy[dot]bnaya[At]gmail[dot]com wrote:

How can this be? can the debugger "lie"?
Hope my question is clearer now..
How is perm initialised. If perm contains all zeros, digit will remain
at its initial condition. Also perm[j] will never be less than perm[i],
as they'll be equal. If Hash starts out at zero it will remain their
under these circumstances. If we don't see how they're initialised, we
would not be able to comment. On the other hand, that would be doing
your work for you :-0.

W Thanks.

[Ben Pope]

zahy[dot]bnaya[At]gmail[dot]com wrote:

Again, I did not write this code and the code is doing what it should.
I just want to revise it.
I was expecting the behavior exactly as described by werasm,
But when running it from the debugger it looks like the line:
hashval = hashval * (SIZE - i-1) + digit;
is being executed as part of for2.
However the hashval variable remains at it's intial value (0) although
digit is changing . It looks like the line is not being
executed at all.
How can this be? can the debugger "lie"?
Hope my question is clearer now..

You still haven't posted complete code.

I have seen debuggers become confused, and I have seen debuggers
attempting to show you your original code whilst debugging optimised
code. This usually confuses the user.

We can't help you very well, unless you post compilable code that
demonstrates the problem.

Ben Pope
--
I'm not just a number. To many, I'm known as a string...