An interesting C problem

[gteddy923]

I had a written test of a company the other day, there's an
interesting problem. it's as below:
Write a function with input parameters i and n, then use only one line
of statement to produce the formatted results as below:
i
i+1
....
n
n-1
....
i

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
The problem is 10 points, if you use one loop, 4 points will be
minused, if you use one other statement, 2 points will be minused.

Can anybody figure out if it's feasible without loosing any points?

Thanks,

[Army1987]

<gt*******@gmail.comha scritto nel messaggio
news:11**********************@p15g2000hsd.googlegr oups.com...

>I had a written test of a company the other day, there's an
interesting problem. it's as below:
Write a function with input parameters i and n, then use only one line
of statement to produce the formatted results as below:
i
i+1
...
n
n-1
...
i

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
The problem is 10 points, if you use one loop, 4 points will be
minused, if you use one other statement, 2 points will be minused.

Can anybody figure out if it's feasible without loosing any points?

Thanks,

#include <stdio.h>
void idiotic_thing(int i, int n);
void idiotic_thing(int i, int n)
{
idiotic_thing1(i, n);
}

void idiotic_thing1(int i, int n)
{
int j;
if (i < n) return;
for (j = i; j <= n; j++)
printf("%d\n", j);
for (j = n-1; j >= i; j++)
printf("%d\n", j);
return;
}

The function idiotic_thing itself uses only one statement and no loop.

[santosh]

Army1987 wrote:

<gt*******@gmail.comha scritto nel messaggio
news:11**********************@p15g2000hsd.googlegr oups.com...

I had a written test of a company the other day, there's an
interesting problem. it's as below:
Write a function with input parameters i and n, then use only one line
of statement to produce the formatted results as below:
i
i+1
...
n
n-1
...
i

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
The problem is 10 points, if you use one loop, 4 points will be
minused, if you use one other statement, 2 points will be minused.

Can anybody figure out if it's feasible without loosing any points?

Thanks,

#include <stdio.h>
void idiotic_thing(int i, int n);

ITYM, void idiotic_thing1(int i, int n);

void idiotic_thing(int i, int n)
{
idiotic_thing1(i, n);
}

void idiotic_thing1(int i, int n)
{
int j;
if (i < n) return;

This check squashes the purpose of this routine.

for (j = i; j <= n; j++)
printf("%d\n", j);
for (j = n-1; j >= i; j++)
printf("%d\n", j);
return;
}

The function idiotic_thing itself uses only one statement and no loop.

[Richard Heathfield]

gt*******@gmail.com said:

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
The problem is 10 points, if you use one loop, 4 points will be
minused, if you use one other statement, 2 points will be minused.

Can anybody figure out if it's feasible without loosing any points?

A loop is an iteration-statement, which is one of the several different
kinds of statement defined by the grammar, so I see no reason why it
should lose any points. When your questioner can produce this:

*
* *
* *
* *
* *
* *
*

in a single bodyless iteration-statement (where the height of the
diamond is configurable at runtime), I'll think about thinking about
his problem. Until then, just tell him he's behind the curve.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Malcolm McLean]

<gt*******@gmail.comwrote in message

>I had a written test of a company the other day, there's an
interesting problem. it's as below:
Write a function with input parameters i and n, then use only one line
of statement to produce the formatted results as below:
i
i+1
...
n
n-1
...
i

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
The problem is 10 points, if you use one loop, 4 points will be
minused, if you use one other statement, 2 points will be minused.

Can anybody figure out if it's feasible without loosing any points?

Think recursion.
--
Free games and programming goodies.
http://www.personal.leeds.ac.uk/~bgy1mm

[Lauri Alanko]

In article <11**********************@p15g2000hsd.googlegroups .com>,
gt*******@gmail.com <gt*******@gmail.comwrote:

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.

Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.
Lauri

[Army1987]

"santosh" <sa*********@gmail.comha scritto nel messaggio
news:11**********************@o5g2000hsb.googlegro ups.com...

>void idiotic_thing1(int i, int n)
{
int j;
if (i < n) return;

This check squashes the purpose of this routine.

Whoops... I meant (i n) (but I think it's unnecessary anyway...)

[Army1987]

"Lauri Alanko" <[e-mail] scritto nel messaggio
news:eu**********@oravannahka.helsinki.fi...

Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Also, think about the behaviour of && and || when the second operand has
side effects.

[Barry]

"Army1987" <pl********@for.itwrote in message
news:eu**********@tdi.cu.mi.it...

>
"Lauri Alanko" <[e-mail] scritto nel messaggio
news:eu**********@oravannahka.helsinki.fi...

>Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Also, think about the behaviour of && and || when the second operand has
side effects.

No && or || are required.

[Army1987]

"Barry" <ba****@nullhighstream.netha scritto nel messaggio
news:13*************@corp.supernews.com...

>Also, think about the behaviour of && and || when the second operand has
side effects.

No && or || are required.

There is more than one way to do that. I used && and ||, but used no ternary
operator.

[Bill Pursell]

On Apr 1, 2:46 pm, "Army1987" <[e-mail]:

"Barry" <[e-mail] scritto nel messaggionews:13*************@corp.supernews.com.. .

Also, think about the behaviour of && and || when the second operand has
side effects.

No && or || are required.

There is more than one way to do that. I used && and ||, but used no ternary
operator.

But did you check the return value of printf and handle
errors gracefully?

[army1987]

On 1 Apr, 22:06, "Bill Pursell" <[e-mail]:

There is more than one way to do that. I used && and ||, but used no ternary
operator.

But did you check the return value of printf and handle
errors gracefully?

Ehm... no, but if that matters, you can include the right headers, set
errno to 0, call the function, and then check errno...

Anyway, I never used printf as the left operator of &&, if you meant
*that*. In the function I wrote, it is always used as the left
"operand" of a comma.

[CBFalconer]

Barry wrote:

"Army1987" <pl********@for.itwrote in message

>"Lauri Alanko" <[e-mail] scritto nel messaggio

>>Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then
figure out how to combine them to get the desired result.

Also, think about the behaviour of && and || when the second
operand has side effects.

No && or || are required.

Neither is the comma operator nor printf (for n < 10). putc will
suffice.

void doit(int n) {
if (n) {
putc(n + '0'); putc('\n');
doit(n - 1);
if (n 1) {
putc(n + '0'); putc('\n');
}
}
}

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[Keith Thompson]

ar******@email.it writes:

On 1 Apr, 22:06, "Bill Pursell" <[e-mail]:

There is more than one way to do that. I used && and ||, but used
no ternary operator.

But did you check the return value of printf and handle
errors gracefully?

Ehm... no, but if that matters, you can include the right headers, set
errno to 0, call the function, and then check errno...

[...]

Or, better yet, you can check the value returned by printf. The standard
doesn't say that printf set error on errors.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

Keith Thompson <ks***@mib.orgwrites:

ar******@email.it writes:

>On 1 Apr, 22:06, "Bill Pursell" <[e-mail]:

>There is more than one way to do that. I used && and ||, but used
no ternary operator.

But did you check the return value of printf and handle
errors gracefully?

Ehm... no, but if that matters, you can include the right headers, set
errno to 0, call the function, and then check errno...

[...]

Or, better yet, you can check the value returned by printf. The standard
doesn't say that printf set error on errors.

Whoops, I meant that the standard doesn't say that printf sets *errno*
on errors.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[gteddy923]

On Apr 1, 8:14 pm, Lauri Alanko <[e-mail]:

In article <[e-mail] .com>,

[e-mail] <[e-mail]:

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.

Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Lauri

Thanks, i think i've figured out how to do that.
int func(int i, int n)
{

return printf("%d\n", i) && i<n && ( p(i + 1, n) || 1) && printf("%d
\n", i);
}

The func works as required :)

[Laurent Deniau]

gt*******@gmail.com wrote:

On Apr 1, 8:14 pm, Lauri Alanko <[e-mail]:

>In article <[e-mail] .com>,

[e-mail] <[e-mail]:

>>For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1
The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.

Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Lauri

Thanks, i think i've figured out how to do that.
int func(int i, int n)
{

return printf("%d\n", i) && i<n && ( p(i + 1, n) || 1) && printf("%d
\n", i);
}

I guess that 'p' must be read as 'func'.

The func works as required :)

But according to the norm, this is not a statement (6.8) but a
function-definition (6.9.1). While

for (int i = 1; i < 2*n; i++)
printf("%d\n", i < n ? i : 2*n-i);

is a statement. And there is many other ways to write it with a single
statement.

a+, ld.

[Laurent Deniau]

Laurent Deniau wrote:

gt*******@gmail.com wrote:

>On Apr 1, 8:14 pm, Lauri Alanko <[e-mail]:

>>In article <[e-mail] .com>,

[e-mail] <[e-mail]:
For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1
The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.
Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Lauri

Thanks, i think i've figured out how to do that.
int func(int i, int n)
{

return printf("%d\n", i) && i<n && ( p(i + 1, n) || 1) &&
printf("%d
\n", i);
}

I guess that 'p' must be read as 'func'.

>The func works as required :)

But according to the norm, this is not a statement (6.8) but a
function-definition (6.9.1). While

for (int i = 1; i < 2*n; i++)
printf("%d\n", i < n ? i : 2*n-i);

Sorry, I hadn't read the first post where you say that 'i' may not
necessary start at 1. So here a better version:

for (int j = 2*n-i; i <= j; i++)
printf("%d\n", i < n ? i : 2*n-i);

is a statement. And there is many other ways to write it with a single
statement.

a+, ld.

[Chris Dollin]

Laurent Deniau wrote:

>>>[e-mail] <[e-mail]:

>>>>The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.

Sorry, I hadn't read the first post where you say that 'i' may not
necessary start at 1. So here a better version:

for (int j = 2*n-i; i <= j; i++)
printf("%d\n", i < n ? i : 2*n-i);

>is a statement.

It's a statement, but it isn't ONE statement: I see two.

--
Yes, Virginia, there is a second Jena user conference: Palo Alto, Sep 2007.
"Go not to the Drazi for counsel, Unsaid /Babylon 5/
for they will answer both 'green' and 'purple'."

Hewlett-Packard Limited Cain Road, Bracknell, registered no:
registered office: Berks RG12 1HN 690597 England

[Barry]

<gt*******@gmail.comwrote in message
news:11**********************@o5g2000hsb.googlegro ups.com...

On Apr 1, 8:14 pm, Lauri Alanko <[e-mail]:

>In article <[e-mail] .com>,

[e-mail] <[e-mail]:

For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

The trickest thing is that you should USE ONLY ONE STATEMENT to
accomplish that.

Here are the main ingredients:

expression statement
comma operator
conditional expression
function call expression
the standard function 'printf'

If you aren't familiar with some of them, look them up. Then figure
out how to combine them to get the desired result.

Lauri

Thanks, i think i've figured out how to do that.
int func(int i, int n)
{

return printf("%d\n", i) && i<n && ( p(i + 1, n) || 1) && printf("%d
\n", i);
}

The func works as required :)

When I replied no && or || was required I was thinking:

void doitr(int i, int n)
{
i<n?printf("%d\n",i), doitr(i+1,n), printf("%d\n",i):printf("%d\n",i);
}

[Laurent Deniau]

Chris Dollin wrote:

Laurent Deniau wrote:

>>>>[e-mail] <[e-mail]:

>>>>>The trickest thing is that you should USE ONLY ONE STATEMENT to
>accomplish that.

>Sorry, I hadn't read the first post where you say that 'i' may not
necessary start at 1. So here a better version:

for (int j = 2*n-i; i <= j; i++)
printf("%d\n", i < n ? i : 2*n-i);

>>is a statement.

It's a statement, but it isn't ONE statement: I see two.

It is one statement which includes an iterative-statement which includes
an expression... 'statement' definition in the norm is recursive. If set
A includes set B, it doesn't mean that A is two sets ;-)

a+, ld.

[CBFalconer]

Chris Dollin wrote:

Laurent Deniau wrote:

.... snip ...

>

> for (int j = 2*n-i; i <= j; i++)
printf("%d\n", i < n ? i : 2*n-i);

is a statement.

It's a statement, but it isn't ONE statement: I see two.

Picking miniscule nits, it IS one statement, a 'for' statement.
Yes, that statement is compounded of various other statements, yet
it remains a single statement. It's execution can be controlled by
a single preceding if (clause), etc. After which the if () etc.
would form a single statement.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[Chris Dollin]

CBFalconer wrote:

Chris Dollin wrote:

>Laurent Deniau wrote:

... snip ...

>>

>> for (int j = 2*n-i; i <= j; i++)
printf("%d\n", i < n ? i : 2*n-i);

is a statement.

It's a statement, but it isn't ONE statement: I see two.

Picking miniscule nits, it IS one statement, a 'for' statement.
Yes, that statement is compounded of various other statements, yet
it remains a single statement.

The `for` statement is one statement. But there is another statement
there -- the expression-statement calling printf. (If it were not a
statement, then the `for` would be illegal.)

I see two statements, one of which happens to be a component of the
other.

--
The second Jena user conference! http://hpl.hp.com/conferences/juc2007/
Nit-picking is best done among friends.

Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN

[Army1987]

Neither is the comma operator nor printf (for n < 10). putc will

suffice.

void doit(int n) {
if (n) {
putc(n + '0'); putc('\n');
doit(n - 1);
if (n 1) {
putc(n + '0'); putc('\n');
}
}
}

You meant putchar(), or putc(..., stdin)?
And that is not *one* statement.

[Army1987]

"Army1987" <pl********@for.itha scritto nel messaggio
news:ev**********@tdi.cu.mi.it...

>Neither is the comma operator nor printf (for n < 10). putc will
suffice.

void doit(int n) {
if (n) {
putc(n + '0'); putc('\n');
doit(n - 1);
if (n 1) {
putc(n + '0'); putc('\n');
}
}
}

You meant putchar(), or putc(..., stdin)?
And that is not *one* statement.

It is one statement. Sorry...

[Army1987]

Thanks, i think i've figured out how to do that.
int func(int i, int n)
{

return printf("%d\n", i) && i<n && ( p(i + 1, n) || 1) && printf("%d
\n", i);
}

If i is 4 and n is 3, it does print 4.
Is it what you meant?

(The one I wrote prints nothing if n < i).

[Army1987]

"Army1987" <pl********@for.itha scritto nel messaggio
news:ev**********@tdi.cu.mi.it...

>Neither is the comma operator nor printf (for n < 10). putc will
suffice.

void doit(int n) {
if (n) {
putc(n + '0'); putc('\n');
doit(n - 1);
if (n 1) {
putc(n + '0'); putc('\n');
}
}
}

You meant putchar(), or putc(..., stdin)?
And that is not *one* statement.

Now I've read the other posts in this thread. It is one statement, but there
are several statements in it. (Nested statements are statements, too. So
this isn't what everybody would call "a function using one statement".
<OT>When I was about 8, once I drew a square divided in four, and asked
several people "How many squares can you see?". Most people anwswered "four"
(the small ones). Almost nobody answered "one". The answer I expected was
"five" (the small ones and the big one).</OT>)

Here is why *this* meaning of "one statement" sounded brain-dead to me.
With this, I could use if (1) to make any arbitrarily complex function "use
one statement". What I think on reading "function using one statement" is a
function with one semicolon in its body.

[CBFalconer]

Army1987 wrote:

>

>Neither is the comma operator nor printf (for n < 10). putc will
suffice.

void doit(int n) {
if (n) {
putc(n + '0'); putc('\n');
doit(n - 1);
if (n 1) {
putc(n + '0'); putc('\n');
}
}
}

You meant putchar(), or putc(..., stdin)?
And that is not *one* statement.

Yes, should be putchar. Nothing in the article to which I replied
said anything about one statement. The problem was to print:

N
N-1
....
1
2
....
N

--
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<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>

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-- Francis Crick, co-discover of DNA
"There is nothing more amazing than stupidity in action."
-- Thomas Matthews

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[Old Wolf]

On Apr 1, 10:31 pm, "[e-mail]" <[e-mail]:

Write a function with input parameters i and n, then use only one line
of statement to produce the formatted results as below:
For example, let i = 1, n = 4, then the output should be:
1
2
3
4
3
2
1

void func(int i, int n)
{
printf("%d\n", i), i < n && (func(i + 1, n), printf("%d\n", i));
}

Least inelegant correct answer so far?