Puzzle: Add One Without Using + OR -

[Peter Koch Larsen]

But the least thought provoking too...

Not at all. It is the solution that understand how to take advantage of C++.
Most of the other solutions are simply C bit-fiddling, while some use C++
but in a most inefficient way (Jonathans std::vector).

/Peter

 

[Alf P. Steinbach]

* Dietmar Kuehl:

* DaKoadMunky:


I wouldn't ask for such a thing but if it became apparent that this
question could not be answered, I would probably indeed deny a C++
job. After all, the simple solution 'std:lus<int>()(n, 1)' should be
obvious to anybody knowing the standard C++ library... Funny that I saw
it mentioned only once before in this thread: this was first solution I
had in mind.

Probably because the question was C/C++ program, not C++ program (but I may be
wrong, just judging by my own response).

 

[Jyrki Alakuijala]

#include <math.h>

void addOne(unsigned int &a) {
a = log(4 * exp(a));
}

Please, do remember to have sufficient exponent accuracy in
your floating point representation to cover the whole unsigned
int range.

 

[Jyrki Alakuijala]

Adding without loops:

void addOne(unsigned int &val) {
unsigned int x = ~val;
#define A(y) x &= (x & y) ? y : ~y;
A(0x0000ffff)
A(0x00ff00ff)
A(0x0f0f0f0f)
A(0x33333333)
A(0x55555555)
#undef A
val &= x * ~0;
val |= x;
}

 

[Matt Hurd]

Excellent! This one is the cleanest and best.

Given he didn't say which unsigned integer (overly pedantic hat) you
could go for some extra credit ;-)

#include <functional>

template< class T >
T add_one(T x)
{
return std:lus<T>()(x, 1);
}

matt.
www.hurd.com.au

 

[Gernot Frisch]

Adding without loops:

void addOne(unsigned int &val) {
unsigned int x = ~val;
#define A(y) x &= (x & y) ? y : ~y;
A(0x0000ffff)
A(0x00ff00ff)
A(0x0f0f0f0f)
A(0x33333333)
A(0x55555555)
#undef A
val &= x * ~0;
val |= x;
}

This one wins the obsfruscation contest...