How to make an adder function with these characters?

Question

I saw a "puzzle", where you have to write a function in C that returns the value of a+c, however, you can't use the + operator.

unsigned f(unsigned a, unsigned c) {
    return <write your coe here>;
}

You can use only the following characters: harc()&|[]*/.

Looks like they want you to invoke undefined behavior by casting an input to a `char *` and using subscripting for pointer arithmetic. **This is not guaranteed to work.** — user2357112, Jan 02 '16 at 13:15
@EdHeal: Is that a good hint? I don't see how that would work. — Karoly Horvath, Jan 02 '16 at 13:19
Is it necessary to use all characters in `harc()&|[]*/` or lesser can be used? — haccks, Jan 02 '16 at 13:22
@klenium; Good. Are you sure last character in the list is `/` and not `\\` ? — haccks, Jan 02 '16 at 13:23
@haccks - doesn't matter - you don't need it. There's one other you don't need either. — enhzflep, Jan 02 '16 at 13:24
This is close to what you seem to be looking for: http://stackoverflow.com/a/366247/1187415. — Martin R, Jan 02 '16 at 13:28

score 5 · Accepted Answer · edited Jan 02 '16 at 15:17

5

This should work in practice, though I believe it relies on undefined behaviour and will generate warnings:

unsigned f(unsigned a, unsigned c) {
    return &(((char*)a)[c]);
}

(fewer brackets are necessary in practice)

This works as follows:

   (char*)a      - cast 'a' to a 'char *' pointer 
  ((char*)a)[c]  - treat 'a' as an array, and index the c'th element, which
                   will be at address c + a
&(((char*)a)[c]) - take the address of that element, i.e. c + a

Finally the return casts this back to unsigned.

Trivial test harness which compiles with gcc 4.8 with two warnings:

#include <stdio.h>

unsigned
f (unsigned a, unsigned c)
{
  return &(((char *) a)[c]);
}

int
main (int argc, char **argv)
{
  unsigned a = 1;
  unsigned b = 2;
  unsigned c = f (a, b);
  printf ("%d + %d = %d\n", a, b, c);
  return 0;
}

Note as this is C, not C++, it may not compile with a C++ compiler.

edited Jan 02 '16 at 15:17

Crowman

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answered Jan 02 '16 at 13:25

abligh

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1

Awesome! And can you explain why it works/what's happening? – klenium Jan 02 '16 at 13:30
2

`error: invalid conversion from 'char*' to 'unsigned int'` is what I get – Ziezi Jan 02 '16 at 13:31
@simplicisveritatis - g++ 4.7.something vomits this error message, while gcc 4.7.sameSomething turns it into working code. – enhzflep Jan 02 '16 at 13:35
1

@simplicisveritatis Compiles file here with gcc 4.8 (albeit with two warnings) and works. Compiling with g++ won't work as g++ compiles C++ and this is C. – abligh Jan 02 '16 at 13:36
Just cured it using `reinterpret_cast` if you want add it to your answer. Ugly, but it works ! :) – Ziezi Jan 02 '16 at 13:38
@simplicisveritatis no, that would not be OK as most of the characters in `reinterpret_cast` are not permitted. This is a C challenge, not a C++ challenge, and `reinterpret_cast` is not valid C. C and C++ are different (albeit related) languages. – abligh Jan 02 '16 at 13:39
@simplicisveritatis, no, that would not be OK either as none of the characters `i`, `n` or `t` are permitted. – abligh Jan 02 '16 at 13:41
actually, the posted answer compiles with 4 warnings, 1) unused parameter `argc` 2) unused parameter `argv` 3) line 6 cast to pointer to integer of different size 4) return makes integer from pointer without a cast – user3629249 Jan 04 '16 at 06:39
suggest changing the signature of the main() function to: `int main( void )` – user3629249 Jan 04 '16 at 06:40

How to make an adder function with these characters?

1 Answers1