# A for loop iterating over the complete range of a variable

Discussion in 'C Programming' started by Joona I Palaste, Aug 29, 2004.

1. ### Joona I PalasteGuest

Reading Skybuck Flying's obvious troll message, I thought of how to
properly do a for loop that would iterate over the complete range of
an unsigned variable. As you all know, Skybuck's method won't work.
There are a number of other ways. You could use a wider type to do the
counting, but that would be cheating. You could also handle either the
first or last iteration as a special case outside the loop, but then it
would be more than a loop. You could make a very big array of integer
flags to tell whether you've already visited a value, but that would
consume too much memory.
So I settled down to this version. Assume unsigned short is 16 bits
wide.

int stop = 0;
unsigned short i;
for (i=0; !stop; stop = ++i==0) {
/* do something */
}

Any more elegant solutions out there?

--
/-- Joona Palaste () ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"Ice cream sales somehow cause drownings: both happen in summer."
- Antti Voipio & Arto Wikla

Joona I Palaste, Aug 29, 2004

2. ### Joona I PalasteGuest

Joona I Palaste <> scribbled the following:
> So I settled down to this version. Assume unsigned short is 16 bits
> wide.

Actually you don't need to assume that...

> int stop = 0;
> unsigned short i;
> for (i=0; !stop; stop = ++i==0) {
> /* do something */
> }

> Any more elegant solutions out there?

--
/-- Joona Palaste () ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"I wish someone we knew would die so we could leave them flowers."
- A 6-year-old girl, upon seeing flowers in a cemetery

Joona I Palaste, Aug 29, 2004

3. ### Arthur J. O'DwyerGuest

On Sun, 29 Aug 2004, Joona I Palaste wrote:
>
> Reading Skybuck Flying's obvious troll message, I thought of how to
> properly do a for loop that would iterate over the complete range of
> an unsigned variable. As you all know, Skybuck's method won't work.
> There are a number of other ways. You could use a wider type to do the
> counting, but that would be cheating. You could also handle either the
> first or last iteration as a special case outside the loop, but then it
> would be more than a loop. You could make a very big array of integer
> flags to tell whether you've already visited a value, but that would
> consume too much memory.
> So I settled down to this version. Assume unsigned short is 16 bits
> wide.
>
> int stop = 0;
> unsigned short i;
> for (i=0; !stop; stop = ++i==0) {
> /* do something */
> }
>
> Any more elegant solutions out there?

I'm sure you've seen the solution that gets posted every time someone
claims it's hard to do:

unsigned short i = 0;
do {
/* do something */
} while (++i != 0);

It's not a 'for' loop, but it works and is easy to remember and write.
I guess you could use a 'for' loop similar to yours, but it would still
be slower and use a temporary variable: [UNTESTED CODE]

unsigned short i;
int tmp;
for (tmp=0, i=0; i || !tmp; ++i, tmp=1) {
/* do something */
}

-Arthur

Arthur J. O'Dwyer, Aug 29, 2004
4. ### Richard TobinGuest

In article <cgt24f\$9um\$>,
Joona I Palaste <> wrote:

>int stop = 0;
>unsigned short i;
>for (i=0; !stop; stop = ++i==0) {
> /* do something */
>}

If you use a for loop, you are bound to need some extra variable, since
the test at the start has to succeed 65536 times and fail once, so you
need 65537 states.

You can avoid the use of an extra variable by using a do ... while
loop, which tests at the end and therefore needs to succeed only 65535
times:

i = 0;
do {
/* do something */
i++;
} while(i != 0);

-- Richard

Richard Tobin, Aug 29, 2004
5. ### Ed MortonGuest

Joona I Palaste wrote:
> Reading Skybuck Flying's obvious troll message, I thought of how to
> properly do a for loop that would iterate over the complete range of
> an unsigned variable. As you all know, Skybuck's method won't work.
> There are a number of other ways. You could use a wider type to do the
> counting, but that would be cheating. You could also handle either the
> first or last iteration as a special case outside the loop, but then it
> would be more than a loop. You could make a very big array of integer
> flags to tell whether you've already visited a value, but that would
> consume too much memory.
> So I settled down to this version. Assume unsigned short is 16 bits
> wide.
>
> int stop = 0;
> unsigned short i;
> for (i=0; !stop; stop = ++i==0) {
> /* do something */
> }
>
> Any more elegant solutions out there?
>

unsigned short i;
for (i=0; ++i {
/* do something */
}

For the general case where i might be getting incremented by some number
other than 1 and so won't necessarily have the value zero when it loops
around (e.g. 13 in the following example), this might be better:

unsigned short i, j;
for (i=0, j=0; i >= j; j=i, i+=13) {
/* do something */
}

Regards,

Ed.

Ed Morton, Aug 29, 2004
6. ### Christian BauGuest

In article <cgt24f\$9um\$>,
Joona I Palaste <> wrote:

> Reading Skybuck Flying's obvious troll message, I thought of how to
> properly do a for loop that would iterate over the complete range of
> an unsigned variable. As you all know, Skybuck's method won't work.
> There are a number of other ways. You could use a wider type to do the
> counting, but that would be cheating. You could also handle either the
> first or last iteration as a special case outside the loop, but then it
> would be more than a loop. You could make a very big array of integer
> flags to tell whether you've already visited a value, but that would
> consume too much memory.
> So I settled down to this version. Assume unsigned short is 16 bits
> wide.
>
> int stop = 0;
> unsigned short i;
> for (i=0; !stop; stop = ++i==0) {
> /* do something */
> }
>
> Any more elegant solutions out there?

i = 0;
do {
<statements>
++i;
} while (i != 0);

Doing the same for signed int in a portable way seems to be difficult.

Christian Bau, Aug 29, 2004
7. ### CBFalconerGuest

Joona I Palaste wrote:
>
> Reading Skybuck Flying's obvious troll message, I thought of how to
> properly do a for loop that would iterate over the complete range of
> an unsigned variable. As you all know, Skybuck's method won't work.
> There are a number of other ways. You could use a wider type to do the
> counting, but that would be cheating. You could also handle either the
> first or last iteration as a special case outside the loop, but then it
> would be more than a loop. You could make a very big array of integer
> flags to tell whether you've already visited a value, but that would
> consume too much memory.
> So I settled down to this version. Assume unsigned short is 16 bits
> wide.
>
> int stop = 0;
> unsigned short i;
> for (i=0; !stop; stop = ++i==0) {
> /* do something */
> }
>
> Any more elegant solutions out there?

You need a loop that postpones the test until after the first
iteration. For example:

unsigned int i;

i = -1;
do {
i++;
/* stuff */
} while (i < UINT_MAX);

or

i = 0;
do {
/* stuff */
} while (0U != ++i);

--
A: Because it fouls the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

CBFalconer, Aug 29, 2004
8. ### Keith ThompsonGuest

Christian Bau <> writes:
[...]
> i = 0;
> do {
> <statements>
> ++i;
> } while (i != 0);
>
> Doing the same for signed int in a portable way seems to be difficult.

int i = INT_MIN;
while (1) {
<statements>
if (i == INT_MAX) break;
i ++;
}

(Don't try this at home; depending on how fast your machine is,
iterating over a 32-bit int type could take several minutes and annoy
other users.)

--
Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

Keith Thompson, Aug 29, 2004
9. ### Ben PfaffGuest

Keith Thompson <> writes:

> (Don't try this at home; depending on how fast your machine is,
> iterating over a 32-bit int type could take several minutes and annoy
> other users.)

Most home machines are single-user
--
"Some people *are* arrogant, and others read the FAQ."
--Chris Dollin

Ben Pfaff, Aug 29, 2004
10. ### Tim RentschGuest

Joona I Palaste <> writes:

> Reading Skybuck Flying's obvious troll message, I thought of how to
> properly do a for loop that would iterate over the complete range of
> an unsigned variable.
>
> [snip]
>
> Any more elegant solutions out there?

The code

if( i = LOWER_BOUND, i <= UPPER_BOUND ) do {

/* stuff to do */

} while( i++ != UPPER_BOUND );

does the loop body once for each value of 'i' in the closed
interval [ LOWER_BOUND .. UPPER_BOUND ] whatever the bounds are (*),
and doesn't rely on any wraparound tricks. Any decent compiler
would optimize away the initial test if LOWER_BOUND and UPPER_BOUND
were known at compile time.

(*) As long as they are inside the range of the type of 'i'.
Being outside the range should also should be detected by
any reasonably decent compiler.

Tim Rentsch, Aug 29, 2004
11. ### peteGuest

Arthur J. O'Dwyer wrote:
>
> On Sun, 29 Aug 2004, Joona I Palaste wrote:

> > unsigned short i;
> > for (i=0; !stop; stop = ++i==0) {

> > Any more elegant solutions out there?

> unsigned short i = 0;

> } while (++i != 0);

Those ain't elegant.
++i;
means the same thing as
i = i + 1;

If INT_MAX equals USHRT_MAX, as it may,
then you have undefined behavior when i has the value of INT_MAX
and you try to increment it.

--
pete

pete, Aug 30, 2004
12. ### Arthur J. O'DwyerGuest

On Mon, 30 Aug 2004, pete wrote:
>
> Arthur J. O'Dwyer wrote:
>
>> unsigned short i = 0;

[...]
>> } while (++i != 0);

>
> Those ain't elegant.
> ++i;
> means the same thing as
> i = i + 1;
>
> If INT_MAX equals USHRT_MAX, as it may,
> then you have undefined behavior when i has the value of INT_MAX
> and you try to increment it.

I guess you're right. What's the fix? Would

} while (i=(unsigned short)i+(unsigned short)1);

fix the UB? (And also, it seems odd that '++i' should mean the same
as 'i=i+1', promotions and all; can I have C&V for that, please?)

-Arthur

Arthur J. O'Dwyer, Aug 30, 2004
13. ### Ben PfaffGuest

"Arthur J. O'Dwyer" <> writes:

> (And also, it seems odd that '++i' should mean the same
> as 'i=i+1', promotions and all; can I have C&V for that, please?)

That one's easy. First, in section 6.5.3.1:

2 The value of the operand of the prefix ++ operator is incremented. The result is the new
value of the operand after incrementation. The expression ++E is equivalent to (E+=1).
See the discussions of additive operators and compound assignment for information on
constraints, types, side effects, and conversions and the effects of operations on pointers.

Then in section 6.5.16.2:

3 A compound assignment of the form E1 op = E2 differs from the simple assignment
expression E1 = E1 op (E2) only in that the lvalue E1 is evaluated only once.

--
"I've been on the wagon now for more than a decade. Not a single goto
in all that time. I just don't need them any more. I don't even use
break or continue now, except on social occasions of course. And I
don't get carried away." --Richard Heathfield

Ben Pfaff, Aug 30, 2004
14. ### peteGuest

Arthur J. O'Dwyer wrote:
>
> On Mon, 30 Aug 2004, pete wrote:
> >
> > Arthur J. O'Dwyer wrote:
> >
> >> unsigned short i = 0;

> [...]
> >> } while (++i != 0);

> >
> > Those ain't elegant.
> > ++i;
> > means the same thing as
> > i = i + 1;
> >
> > If INT_MAX equals USHRT_MAX, as it may,
> > then you have undefined behavior when i has the value of INT_MAX
> > and you try to increment it.

>
> I guess you're right. What's the fix? Would
>
> } while (i=(unsigned short)i+(unsigned short)1);
>
> fix the UB?

No.
(sizeof((short)0 + (short)0)) equals (sizeof(int)).

--
pete

pete, Aug 30, 2004
15. ### Peter NilssonGuest

"Arthur J. O'Dwyer" <> wrote in message
news...
> On Mon, 30 Aug 2004, pete wrote:
> > Arthur J. O'Dwyer wrote:
> >
> > > unsigned short i = 0;

> [...]
> > > } while (++i != 0);

> >
> > Those ain't elegant.
> > ++i;
> > means the same thing as
> > i = i + 1;
> >
> > If INT_MAX equals USHRT_MAX, as it may,
> > then you have undefined behavior when i has the value of INT_MAX
> > and you try to increment it.

>
> I guess you're right. What's the fix? Would
>
> } while (i=(unsigned short)i+(unsigned short)1);
>
> fix the UB?

No. Even though you cast the operands of +, they are still subject to
integral promotion prior to the addition taking place.

The fix is trivial though...

unsigned short i = 0;
do {

} while (i += 1u);

--
Peter

Peter Nilsson, Aug 30, 2004
16. ### xaraxGuest

"Ben Pfaff" <> wrote in message
news:...
> Keith Thompson <> writes:
>
> > (Don't try this at home; depending on how fast your machine is,
> > iterating over a 32-bit int type could take several minutes and annoy
> > other users.)

>
> Most home machines are single-user

Yes, but if he does it, I'll still be annoyed about it.

xarax, Aug 30, 2004
17. ### Fao, SeanGuest

Keith Thompson wrote:
> Christian Bau <> writes:
> [...]
>
>>i = 0;
>>do {
>> <statements>
>> ++i;
>>} while (i != 0);
>>
>>Doing the same for signed int in a portable way seems to be difficult.

>
>
> int i = INT_MIN;
> while (1) {
> <statements>
> if (i == INT_MAX) break;
> i ++;

ITYM i++; ;-)

Sean
--
Sean Fao
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>