m3ph1st0s's programming puzzle 1 (C++) [closed]

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This is the first of a series of C++ puzzles for you. Hope you will enjoy.

So, puzzle no.1:

Given the following program:

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

Insert some code on a single new line anywhere inside the program so that the output will be 0. The new line will contain AT MOST 18 characters (including spaces), and the rest of the lines will remain unmodified. To be clear, here is an example of a valid new code:

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   int* p = NULL;
   const float c=0.5;
   std::cout << a/b-a*c;
}

A new line with 15 characters was inserted so it's ok. However it does not solve the problem.

If this is too simple for you, don't worry, more is coming!!

I'm very happy someone posts a C++ question occasionally! I mean, with all the puzzles where a C++ solution would be 20 or 30 lines, then people posting solutions in J or K or Golfscript becomes frustrating after a while. — Mr Lister, Commented Sep 28, 2012 at 17:59
Does the result have to be valid, well-defined C++ or can it use UB? (But like Mr Lister noted, the original code isn’t even valid C++.) — Konrad Rudolph, Commented Sep 28, 2012 at 18:28
it is not valid. main must return int (read the c++ standard) — BЈовић, Commented Sep 28, 2012 at 19:40
we are missing the point here. put an int and a return 0 if you mind, I didn't. — Bogdan Alexandru, Commented Sep 28, 2012 at 20:00
@Bogdan: Dafuq? For one, DevC++ is so unspeakably ancient, it's output is irrelevant. And secondly, whether or not any given compiler in any given configuration at any time targetting any OS accepts it does not make it valid C++. — DeadMG, Commented Sep 28, 2012 at 20:24

frozenkoi · Accepted Answer · 2012-09-29 02:18:25Z

35

We can get rid of a=1 by moving it into another scope:

#include <iostream>
main() 
{
int a=0;if(0)
    const int a=1;
    const int b=2;
    const float c=0.5;
    std::cout << a/b-a*c;
}

This is I think 13 characters. Or better yet get a new a that also results in 0:

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
if(int a=2)
   std::cout << a/b-a*c;
}

That's 11 characters

answered Sep 29, 2012 at 2:18

frozenkoi

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mob · Accepted Answer · 2012-09-28 15:20:37Z

24

#include <iostream>
main() 
{
   const int a=1;
#define a 0
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

1 new line, 12 new chars

edited Sep 28, 2012 at 15:20

answered Sep 28, 2012 at 14:56

mob

2,60618 silver badges13 bronze badges

\$\begingroup\$ That's what I wanted to submit... +1 for you. \$\endgroup\$
– H2CO3
Commented Oct 1, 2012 at 14:14
\$\begingroup\$ My first idea when read the question))) \$\endgroup\$
– Qwertiy
Commented Jun 25, 2015 at 15:37

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sehe · Accepted Answer · 2012-09-28 20:35:15Z

23

So, #define a 0, Done. I saw that was posted - unsurprisingly.

Surprisingly, this wasn't posted:

#include <iostream>
main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout<<0||
   std::cout << a/b-a*c;
}

14 chars

That should do, right?

answered Sep 28, 2012 at 20:35

sehe

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Mr Lister · Accepted Answer · 2012-09-28 17:56:19Z

22

#include <iostream>
main() 
{
const int a=0;//\
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

17 chars.

By the way, the original program doesn't compile under MSVC, which complains that main doesn't have a return type.

answered Sep 28, 2012 at 17:56

Mr Lister

3,6462 gold badges17 silver badges23 bronze badges

2

\$\begingroup\$ int a=0;//\ will also do the trick \$\endgroup\$
– copy
Commented Sep 28, 2012 at 18:00
2

\$\begingroup\$ Absolutely. But is this a "shortest line wins" kind of contest? \$\endgroup\$
– Mr Lister
Commented Sep 28, 2012 at 18:02
\$\begingroup\$ I thought so, but it's not. Nice trick by the way \$\endgroup\$
– copy
Commented Sep 28, 2012 at 18:10
2

\$\begingroup\$ Yes, shortest solution wins on codegolf SE. see faq \$\endgroup\$
– BЈовић
Commented Sep 28, 2012 at 19:38
2

\$\begingroup\$ @BЈовић the FAQ says the shortest solution wins for actual code-golf questions. This question is not clearly a code-golf question. \$\endgroup\$
– kojiro
Commented Sep 29, 2012 at 14:04

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Dan · Accepted Answer · 2012-09-28 15:26:40Z

14

#define int float

should work as well and is the same length.

answered Sep 28, 2012 at 15:26

Dan

2561 silver badge4 bronze badges

\$\begingroup\$ this is what I had in mind when I first came with the idea \$\endgroup\$
– Bogdan Alexandru
Commented Sep 28, 2012 at 19:41
6

\$\begingroup\$ #define int float is actually undefined behavior. You are not allowed to give keywords new meaning. \$\endgroup\$
– fredoverflow
Commented Sep 28, 2012 at 20:03
\$\begingroup\$ Fred, can you cite your sources? The GCC cpp docs say "You may define any valid identifier as a macro, even if it is a C keyword." \$\endgroup\$
– Dan
Commented Sep 28, 2012 at 20:07
\$\begingroup\$ @Dan: The C++ Standard forbids it. \$\endgroup\$
– DeadMG
Commented Sep 28, 2012 at 20:16
3

\$\begingroup\$ It may be UB, but several popular compilers support it anyway. I've done #define int ERROR to force myself to use the equivalent of int32_t instead of built-in types. By the time I got around to int main(), I had forgotten about the macro and wondered why the heck my code wouldn't compile. \$\endgroup\$
– dan04
Commented Jun 21, 2014 at 1:26

| Show 5 more comments

shiona · Accepted Answer · 2012-09-28 15:02:30Z

12

18, including newline

#define float int

answered Sep 28, 2012 at 15:02

shiona

3,00917 silver badges22 bronze badges

\$\begingroup\$ I'm not really sure this works...ur declaring c as a float and initializing it with 0.5 \$\endgroup\$
– Bogdan Alexandru
Commented Sep 28, 2012 at 19:40
5

\$\begingroup\$ Which will truncate to zero. \$\endgroup\$
– DeadMG
Commented Sep 28, 2012 at 20:15
\$\begingroup\$ you're right, funny thing I've never come across initializing an int with decimal value, I'd thought it would be compiler error, but it only issues a warning \$\endgroup\$
– Bogdan Alexandru
Commented Sep 28, 2012 at 21:31
2

\$\begingroup\$ @BogdanAlexandru take a gander at the C++ standard, it specifically details the implicit conversion at play here. \$\endgroup\$
– obataku
Commented Sep 28, 2012 at 21:39
3

\$\begingroup\$ Note that although all compilers allow this, the standard prohibits redefining keywords (and float is a keyword). \$\endgroup\$
– avakar
Commented Sep 29, 2012 at 6:51

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Andrey Regentov · Accepted Answer · 2013-01-16 05:34:18Z

11

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
   1?std::cout<<0:
   std::cout << a/b-a*c;
}

15 chars.

edited Jan 16, 2013 at 5:34

answered Oct 1, 2012 at 8:34

Andrey Regentov

2111 silver badge4 bronze badges

\$\begingroup\$ will output more than a "0" \$\endgroup\$
– Bogdan Alexandru
Commented Oct 1, 2012 at 12:20
2

\$\begingroup\$ why the ternary operator wouldn't work? \$\endgroup\$
– Andrey Regentov
Commented Oct 11, 2012 at 5:34

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FireFly · Accepted Answer · 2012-09-28 18:41:12Z

9

#include <iostream>
main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
#define a 0;1
   std::cout << a/b-a*c;
}

14 characters.

answered Sep 28, 2012 at 18:41

FireFly

7,28829 silver badges41 bronze badges

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Ashrr · Accepted Answer · 2012-09-29 09:25:51Z

8

#include <iostream>
main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
   return puts("0");
   std::cout << a/b-a*c;
}

17 chars.

answered Sep 29, 2012 at 9:25

Ashrr

812 bronze badges

1

\$\begingroup\$ Best solution yet, because it flies in the face of anything related to C++. \$\endgroup\$
– fabspro
Commented Sep 29, 2012 at 12:10
3

\$\begingroup\$ `puts´ was not declared in this scope \$\endgroup\$
– shiona
Commented Sep 30, 2012 at 7:44
\$\begingroup\$ @shiona what compiler are you using? \$\endgroup\$
– Ashrr
Commented Oct 2, 2012 at 7:16
\$\begingroup\$ @Ashrr gcc (g++) 4.5.4 \$\endgroup\$
– shiona
Commented Oct 2, 2012 at 16:32
1

\$\begingroup\$ Don't think, this code returns 0. But you can replace space via exclamation mark. Anyway, I'm not sure when puts returns zero. \$\endgroup\$
– Qwertiy
Commented Jun 25, 2015 at 15:33

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Roddy of the Frozen Peas · Accepted Answer · 2012-09-29 10:15:28Z

8

#include <iostream>
main()
{
   const int a=1;
   const int b=2;
   const float c=0.5;
std::cout<<0;//\
   std::cout << a/b-a*c;
}

It's 17 characters so it just fits.

edited Sep 29, 2012 at 10:15

Roddy of the Frozen Peas

1031 silver badge3 bronze badges

answered Sep 28, 2012 at 18:33

marinus

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ɐɔıʇǝɥʇuʎs · Accepted Answer · 2014-06-20 21:14:06Z

7

I do not know C++, however based on the question, couldn't you just input a line to simply output 0? the question specifies the output should be 0, it does not specify you must CHANGE the output to 0.

std::cout << 0

(I have 0 clue on C++, perhaps somebody can use this concept though)

edited Jun 20, 2014 at 21:14

ɐɔıʇǝɥʇuʎs

4,65520 silver badges25 bronze badges

answered Sep 28, 2012 at 19:36

NRGdallas

7394 silver badges17 bronze badges

\$\begingroup\$ Oh well, the output should be just 0, I thought it was obvious, otherwise there would be no puzzle, would it? \$\endgroup\$
– Bogdan Alexandru
Commented Sep 28, 2012 at 19:42
8

\$\begingroup\$ @BogdanAlexandru You will find that exploiting a poorly written question is a common technique to solving these puzzles. If you want to prevent users from taking these shortcuts, spend a few extra minutes analyzing your own question and try to remove any potential ambiguity. \$\endgroup\$
– ardnew
Commented Sep 28, 2012 at 22:34

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Csq · Accepted Answer · 2015-06-25 14:24:40Z

3

12 chars, similar to mob's solution

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
#define a b
   std::cout << a/b-a*c;
}

other combinations also work, like #define a c or #define c 0

answered Jun 25, 2015 at 14:24

Csq

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Toby Speight · Accepted Answer · 2015-06-25 15:18:26Z

2

I know it's not code-golf, but I seem to be wearing that hat today!

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;

--a;

   std::cout << a/b-a*c;
}

five chars, including the newline;

answered Jun 25, 2015 at 15:18

Toby Speight

6,5081 gold badge24 silver badges41 bronze badges

2

\$\begingroup\$ This does not compile, because a is const. \$\endgroup\$
– Csq
Commented Jun 25, 2015 at 17:08
\$\begingroup\$ Oops, I should have given it to a compiler! :-( \$\endgroup\$
– Toby Speight
Commented Jun 25, 2015 at 17:20

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baby-rabbit · Accepted Answer · 2012-10-01 02:12:55Z

0

c++ whatever...

echo "0"; exit
#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

run via:

sh mp.cpp

answered Oct 1, 2012 at 2:12

baby-rabbit

1,7339 silver badges11 bronze badges

\$\begingroup\$ Nice try but violates the requirement: valid C++. \$\endgroup\$
– Konrad Rudolph
Commented Oct 1, 2012 at 7:00

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Toby Speight · Accepted Answer · 2015-06-25 15:31:55Z

0

A variant on Mr Lister's answer but slightly less obvious.

#include <iostream>
int main() 
{
   const float a=1; //??/
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

answered Jun 25, 2015 at 15:31

Toby Speight

6,5081 gold badge24 silver badges41 bronze badges

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tow · Accepted Answer · 2012-09-30 17:54:36Z

-1

#include <iostream>
int main() 
{
int a;if(a)
   const int a=1;
   const int b=2;
   const float c=0.5;
   std::cout << a/b-a*c;
}

How about these 11 chars...

answered Sep 30, 2012 at 17:54

tow

101

1

\$\begingroup\$ The problem here is that the int a before the if is not being initialized, so a could have any value. \$\endgroup\$
– frozenkoi
Commented Oct 1, 2012 at 6:35
1

\$\begingroup\$ This is undefined, you’re using an uninitialised value for a. \$\endgroup\$
– Konrad Rudolph
Commented Oct 1, 2012 at 6:59

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Qwertiy · Accepted Answer · 2015-06-25 15:29:57Z

-1

#define a 0

http://codepad.org/N06weGJc

#include <iostream>
int main() 
{
   const int a=1;
   const int b=2;
   const float c=0.5;
#define a 0
   std::cout << a/b-a*c;
}

answered Jun 25, 2015 at 15:29

Qwertiy

3,0091 gold badge12 silver badges31 bronze badges

\$\begingroup\$ Ups.. Posted the answer before looking the others. There is already the same answer: codegolf.stackexchange.com/a/8507/32091 \$\endgroup\$
– Qwertiy
Commented Jun 25, 2015 at 15:37

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m3ph1st0s's programming puzzle 1 (C++) [closed]

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