The Daily WTF: Curious Perversions in Information Technology

Disgruntled DBA · 2006-01-03 Reply Admin

Anonymous:
Otto:
This is true, but the correct response is not to point out x+=11 or even x=x+11, but you ask "Dude, are you high?" and try to score some of that phat dope that Nathan had been smoking.

Because while I can perfectly well see somebody not knowing about += or even zoning out on it, I cannot see anybody, anywhere, ever being unable to grasp a simple x=x+11 statement.

It is obvious that the guy new how to increment with x=x+11, he also rembered that there was an operator similar to x++ but didn't remember it exactly, so he asked his expert colleague. It's just a badly formed question.

No, I think it is a case of a badly formed employee.

jeremydwill · 2006-01-03 Reply Admin

KraGiE:
Gene Wirchenko:
Alex Papadimoulis:

Another slow day, another revisited post. Even if you've seen the original, I highly recommend checking out the comments posted. There you will find a number of solutions (five pages worth) to the problem that Steve Local's ... less gifted ... colleague was having in C# ...

Nathan: Steve, you know how ++ will increment, right?
Steve: Right ....
Nathan: Okay, so how do you increment by 11?

You use +=.

This is not a WTF. It might even be the opposite. Nathan may never have seen += for whatever reason, or it could be a brief zone-out.

Sincerely,

Gene Wirchenko

/agreed. The WTF is more to the OP that clearly isn't able to discern between ignorance and just lack of knowledge. Better he asked than to try to come up with a solution that would have been a WTF.

That is very humorous - "isn't able to discern between ignorance and just lack of knowledge" - neither can I, since they are the same thing! http://dictionary.reference.com/search?q=ignorance

flobi · 2006-01-03 Reply Admin

I submitted that php class I wrote to phpclasses.org to see if they'd actually accept a 150+ line class that just adds 2 numbers together...http://www.phpclasses.org/browse/package/2799.html LOL!

2006-01-03 Reply Admin

Anonymous:
KraGiE:
/agreed. The WTF is more to the OP that clearly isn't able to discern between ignorance and just lack of knowledge. Better he asked than to try to come up with a solution that would have been a WTF.

Even if he's never seen +=. He's probably i = i + 11... Which is the same thing...

Clearly the issue is not that he'd never SEEN the += operator, but that he couldn't REMEMBER the += operator. He knew it existed (as it is essentially the "increment by" operator he's referring to), but didn't recall what it was. Noob behavior, to be sure, but not really "WTF"...

Mystery · 2006-01-03 Reply Admin

#define SIX 8 - 2 #define NINE 2 + 7

x += SIX * NINE;

mgla · 2006-01-04 Reply Admin

Well, of course it has to be:

((((((((((i++)++)++)++)++)++)++)++)++)++)++

[:D]

Gene Wirchenko · 2006-01-04 Reply Admin

Mystery:
#define SIX 8 - 2
#define NINE 2 + 7
x += SIX * NINE;

If you prefer something more canonical:

 #define SIX 1+5
 #define NINE 8+1

 printf("%d", SIX*NINE);

Sincerely,

Gene Wirchenko

brazzy · 2006-01-04 Reply Admin

Gene Wirchenko:
Mystery:
#define SIX 8 - 2
#define NINE 2 + 7
x += SIX * NINE;

If you prefer something more canonical:

 #define SIX 1+5
 #define NINE 8+1

 printf("%d", SIX*NINE);

I'm sure you mean

 #define NINE 8-1

Only then do you get the correct answer (though you may not like it). Even better:

 #define l -2 // correction factor
 ...
 #define NINE 8+l

Gene Wirchenko · 2006-01-04 Reply Admin

brazzy:
Gene Wirchenko:
Mystery:
#define SIX 8 - 2
#define NINE 2 + 7
x += SIX * NINE;

If you prefer something more canonical:

 #define SIX 1+5
 #define NINE 8+1

 printf("%d", SIX*NINE);

I'm sure you mean

 #define NINE 8-1

Only then do you get the correct answer (though you may not like it). Even better:

 #define l -2 // correction factor
 ...
 #define NINE 8+l

Nope. My post stands. Remember to do the substitution first and that there are no brackets.

Sincerely,

Gene Wirchenko

Mystery · 2006-01-04 Reply Admin

Gene Wirchenko:
Mystery:
#define SIX 8 - 2
#define NINE 2 + 7
x += SIX * NINE;
If you prefer something more canonical:

Yeah, I was making reference to that. It still works in my case: forty-two (which equals six times nine) is the answer to the ultimate question of life, the universe, and everything. :-)

Mikademus · 2006-01-04 Reply Admin

mgla:
Well, of course it has to be:

((((((((((i++)++)++)++)++)++)++)++)++)++)++

[:D]

HA! N3wb! </1337> Now go look up the postincrementation rules! And then read MY _informed_ post above where I do it *correctly*! ;) :p

brazzy · 2006-01-05 Reply Admin

Gene Wirchenko:

Nope. My post stands. Remember to do the substitution first and that there are no brackets.

Ah, serves me right for wandering unprepared into the murky depths of C preprocessor madness...

RevMike · 2006-01-09 Reply Admin

No one mentioned:

#!/usr/bin/ruby
i = 1
11.times {i+=1}
puts i

olddog · 2006-01-10 Reply Admin

flobi:
I submitted that php class I wrote to phpclasses.org to see if they'd actually accept a 150+ line class that just adds 2 numbers together...http://www.phpclasses.org/browse/package/2799.html LOL!

Yeah... some poor beg's gonna see it and use it somewhere. Next thing ya know... it'll show up here. But still, what a hoot.

2006-01-11 Reply Admin

What about... x = x + 11.

So damn simple.

2006-01-11 Reply Admin

forty-two (which equals six times nine)

WTF?

brazzy · 2006-01-11 Reply Admin

Anonymous:

forty-two (which equals six times nine)
WTF?

http://en.wikipedia.org/wiki/The_Answer_to_Life%2C_the_Universe%2C_and_Everything

masklinn · 2006-01-11 Reply Admin

Anonymous:

forty-two (which equals six times nine)

WTF?

Traceback (most recent call last): User "Anonymous/Barruelix", line 1, in -toplevel- WTF? GeekKnowledgeException: user doesn't know Douglas Adam

Seriously, in that day and age and on this board, how's that possible?

Maurits · 2006-01-12 Reply Admin

Any technology sufficiently advanced is indistinguishable from a one-line Perl script:
perl -e '$i = shift; eval q($i++;) x 11; print $i' 31

(Windows users -- change the 's to "s)

mgla · 2006-01-20 Reply Admin

Mikademus:
mgla:

Well, of course it has to be:

((((((((((i++)++)++)++)++)++)++)++)++)++)++

[:D]

HA! N3wb! </1337>Now go look up the postincrementation rules! And then read MY _informed_ post above where I do it *correctly*! ;) :p

Thank you for the kind words... [:#]

Did you really think I meant this??? Too bad for you if you did...

salvobeta · 2006-01-26 Reply Admin

I'm surprised no one has shown how easy this is in Ruby!

11.times { i += 1 } i

2006-01-26 Reply Admin

except when i == 0 or (maybe) i < 0

dhromed · 2006-01-27 Reply Admin

The following does not increment by 11:

[code]
var x = 1;

for (var i = 0; i < 4; i++)
{
x += x
}
x -= 5

alert(x)
[/code[

bobo1on1 · 2006-01-27 Reply Admin

int MyNumber;
//assign some value to MyNumber here

union PleaseIncreaseThisIntegerBy11
      {
       signed int number1;
       unsigned int number2;
      };

PleaseIncreaseThisIntegerBy11 PrettyPlease;
PrettyPlease.number2 = 4294967285;
MyNumber -= PrettyPlease.number1;

2006-02-20 Reply Admin

Sorry to bring up an old post, but
maybe (excuse the lack of proper c# but i'm fairly lazy)
sqlConnection sc=new sqlConnection ("blahblah blah server info blah blah");
sc.open();
for (int x=15; x>=5; --x)
new sqlcommand("insert into tablep", sc).executenonquery();
int ineed11=ineed11+int.parse(new sqlcommand("select count (*) from (select top 11 from tablep) q ", sc).executescalar().tostring())

//sc.close() //comment this out because closing connections is for chumps

2006-03-23 Reply Admin

x = ((x/11)++)*11

bobo1on1 · 2006-03-27 Reply Admin

Anonymous:

x = ((x/11)++)*11

You'll need to cast it to a float for small numbers.

2006-03-29 Reply Admin

Here's my addition to the heap of obfuscation seen here (this is in PHP, and was indeed verified to work):

<?
function add11to($i)
{
$i3 = $i; $i2 = $i;
while ($i++ != $i2+count(file(FILE)) - 1){
for($i4 = 0; $i4 <= $i2+count(file(FILE)) - 1; $i4++){
if ($i4 == $i2+count(file(FILE)) - 1) $i3++;
}
}
return $i3;
}
?>

Maurits · 2006-04-06 Reply Admin

bobo1on1:
Anonymous:

x = ((x/11)++)*11

You'll need to cast it to a float for small numbers.

bobo1on1: Huh?

Anonymous: (x/11) is not ++-able. On either side.

2006-04-07 Reply Admin

I was kinda replying to the comment about 'subtle floating point errors' from a previous page, sorry I wasn't more clear, and it should probably have been more like (x/11+1)*11 but I was too lazy to fire up a compiler...

In compensation, here is an example, that if I recall correctly would have worked in GW Basic:

x=x+(x=(x-11))*-11

Maurits · 2006-04-07 Reply Admin

Oh, I see... I was thinking integer arithmetic.

2006-04-07 Reply Admin

i += 11 - i;

Bullet · 2006-04-07 Reply Admin

Alex Papadimoulis:

Another slow day, another revisited post. Even if you've seen the original, I highly recommend checking out the comments posted. There you will find a number of solutions (five pages worth) to the problem that Steve Local's ... less gifted ... colleague was having in C# ...

Nathan: Steve, you know how ++ will increment, right?
Steve: Right ....
Nathan: Okay, so how do you increment by 11?

((var / 11.0)++) * 11.0

Zom-B · 2006-04-17 Reply Admin

dhromed:
I'd like to see some code that almost adds 11.

Preferrably by introducing floating point errors.

double incr = 1.0/3
for (int c = 0; c < 33; c++) i += incr;

1.0/3 == 0.33333333333333331482961625624739
coudn't use the well known 1.0/10 because it would be slightly more (0.10000000000000000555111512312578) instead of less.

2006-05-04 Reply Admin

As you surely know, iteration is human, but recursion is godlike.
Therefore, a much better way to increment i by 11 has to be implemented using recursion:

public class SimpleAdderByEleven {
private int add(int a, int b) throws Exception {
if ((a<0) || (b<0))
throw new Exception("Argument out of range");
if (b>0)
return add(increment(a),decrement(b));
else
return a;
}
public int incrementByEleven (int numberToBeIncrementedByEleven) throws Exception
{
try {
return add(eleven(),numberToBeIncrementedByEleven);
}
catch(Exception e) {
throw new Exception("Argument out of range");
}
}
private int generateEleven(int elevenSeed) {
if (elevenSeed < 11)
return(generateEleven(++elevenSeed));
else
return 11;
}
private int eleven(){
return generateEleven(0);
}
private int increment(int numberToBeIncremented) {
numberToBeIncremented = numberToBeIncremented + (eleven() / eleven());
return numberToBeIncremented;
}
private int decrement(int numberToBeDecremented) {
numberToBeDecremented = numberToBeDecremented - (eleven() / eleven());
return numberToBeDecremented;
}
}

Note: This brilliant piece of code is based on a real homework task (Implement an adder by recursion!) we were assigned at Informatics class at my technical university where I study EE!

2006-05-04 Reply Admin

Sorry, my code somehow was cut off in the middle.
The full version is:
public class SimpleAdderByEleven {
 private int add(int a, int b) throws Exception {
 if ((a<0) || (b<0))
 throw new Exception("Argument out of range");
 if (b>0)
 return add(increment(a),decrement(b));
 else
 return a;
 }
 public int incrementByEleven (int numberToBeIncrementedByEleven) throws Exception
 {
 try {
 return add(eleven(),numberToBeIncrementedByEleven);
 }
 catch(Exception e) {
 throw new Exception("Argument out of range");
 }
 }
 private int generateEleven(int elevenSeed) {
 if (elevenSeed < 11)
 return(generateEleven(++elevenSeed));
 else
 return 11;
 }
 private int eleven(){
 return generateEleven(0);
 }
 private int increment(int numberToBeIncremented) {
 numberToBeIncremented = numberToBeIncremented + (eleven() / eleven());
 return numberToBeIncremented;
 }
 private int decrement(int numberToBeDecremented) {
 numberToBeDecremented = numberToBeDecremented - (eleven() / eleven());
 return numberToBeDecremented;
 }

bakkdoor · 2006-07-06 Reply Admin

omg, i think i found the real solution:

public void AddEleven(ref int value)
{
 int tmp;
 int add = 0;

 List<int> list = new List<int>();

 while(list.Count <= 11)
 {
 list.Add(1);
 }

 add = list.Count;

 tmp = value + add;

 value = tmp;
}

2006-07-06 Reply Admin

int increment_by_eleven(int n) {

double pivot = INT_MAX/2.0;
double x = rand(), old_x = n;

while((x < old_x+10.999) || (x > old_x+11.1)) {
 if(x < old_x+11) x += pivot;
 else x -= pivot;

pivot /= 2.0;
 }
 return (int)x;
}

I think that's as close as anyone can get to the real value, from my tests I'm getting about 99.2% success rate. Who would have thought that binary search would finally pull through.

Some Idiot · 2006-07-06 Reply Admin

Nathan: Steve, you know how ++ will increment, right?
Steve: Right ....
Nathan: Okay, so how do you increment by 11?
Steve: Oh, that's a bit of a problem. At least the first part is easy, the second part's a problem, but not impossible.
Nathan: What's the first part?
Steve: Paste this into notepad and save it. Make sure the first text appears on the first line:

\1+\v
|:-1<
>$@

Nathan: Okay, so what's the second part?
Steve: You're going to have to write a befunge interpreter that allows you to populate the stack before you execute your script. Push the number you need to increment onto the stack, then push 11, then load your script and run it. Once it's finished, you can simply pop your answer right of the stack.
Nathan: Gee, thanks Steve, you're so much more helpful than that CS lecturer I had.

daen · 2006-07-11 Reply Admin

I can't believe nobody has yet come up with the obvious FORTH solution (using prime factors of 42, of course, and a small adjustment)...

: plus11 ( i -- i ) 7 3 2 + + + 1 - ;

Example:

3 plus11 . 14 ok

Tsh. Amateurs.

D.

2006-07-14 Reply Admin

The obvious solution is:

double incby(int x, int inc) { int32_t v2; // change to 64 bit if you have the time double v1; v1 = 0; v2 = inc; do{ while( fabs((v2 *= 16807)/2.e10 + 1.72e-9) > 0.1); v1 += v2/2.e10; //printf("%g\n",v1); =) }while( fabs(v1 - inc) > 1.e0);//change to 1.e-9 for extra fun return x + v1; }

If you really have the time, you can change error and random number generator ... there is no guarantee that this will work for ALL values of inc ^^

bakkdoor · 2006-07-20 Reply Admin

i made up a much easier to read version:

void* addOneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean(void* theValueToWhichWeWillAddOneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean)
{
 int oneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean = 11;
 int temporaryVariableWhichDoesntHaveAnyPurposeButToFillThatUpcomingForLoop = 0;

 for(int one = 1; i <= oneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean; one++)
 {
 temporaryVariableWhichDoesntHaveAnyPurposeButToFillThatUpcomingForLoop++;
 }

 (int)theValueToWhichWeWillAddOneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean += oneMoreThanTenButAtTheSameTimeOneLessThanTwelveIfYouKnowWhatIMean;

 if(thisIsCockyEnough)
 {
 return 0;
 }
}

2006-07-21 Reply Admin

I think that the proper way to approach a situation like this involves the scripting language Lua. This sample generates 20002 lines of Lua script: int f(int n) { lua_State *L = lua_open(); std::ostringstream x; x << "function f(n)\n"; for(int i = -9999; i < 9999; ++i) x << "\tif(n == " << i << ") then return " << i + 11 << " end\n"; x << "error("PANIC")\nend\n"; luaL_loadstring(L, x.str().c_str()); lua_call(L, 0, 0); lua_getglobal(L, "f"); lua_pushnumber(L, n); lua_call(L, 1, 1); int r = lua_tonumber(L, -1); lua_pop(L, 1); // keep the stack clean lua_close(L); }

2006-07-21 Reply Admin

And for advanced programmers:

Just how do you increment 11 by 11 ?

Some Idiot · 2006-07-30 Reply Admin

Anonymous:
And for advanced programmers:

Just how do you increment 11 by 11 ?

((1 << 1
)|1 | ( 1<<
((1 << 1)
| 1)))<<1;

2006-08-03 Reply Admin

i just found the obvious solution!

try{

x+1=x;

}