Elegantly determine if more than one boolean is “t

问题:

I have a set of five boolean values. If more than one of these are true I want to excecute a particular function. What is the most elegant way you can think of that would allow me to check this condition in a single if() statement? Target language is C# but I'm interested in solutions in other languages as well (as long as we're not talking about specific built-in functions).

One interesting option is to store the booleans in a byte, do a right shift and compare with the original byte. Something like if(myByte && (myByte >> 1)) But this would require converting the separate booleans to a byte (via a bitArray?) and that seems a bit (pun intended) clumsy... [edit]Sorry, that should have been if(myByte & (myByte - 1)) [/edit]

Note: This is of course very close to the classical "population count", "sideways addition" or "Hamming weight" programming problem - but not quite the same. I don't need to know how many of the bits are set, only if it is more than one. My hope is that there is a much simpler way to accomplish this.

回答1:

How about

  if ((bool1? 1:0) + (bool2? 1:0) + (bool3? 1:0) + 
      (bool4? 1:0) + (bool5? 1:0) > 1)
      // do something

or a generalized method would be...

   public bool ExceedsThreshold(int threshold, IEnumerable<bool> bools)
    {
       int trueCnt = 0;
       foreach(bool b in bools)
          if (b && (++trueCnt > threshold)) 
              return true;
       return false;          
    } 

or using LINQ as suggested by other answers:

    public bool ExceedsThreshold(int threshold, IEnumerable<bool> bools)
    { return bools.Count(b => b) > threshold; }

EDIT (to add Joel Coehoorn suggestion: (in .Net 2.x and later)

    public void ExceedsThreshold<T>(int threshold, 
                      Action<T> action, T parameter, 
                      IEnumerable<bool> bools)
    { if (ExceedsThreshold(threshold, bools)) action(parameter); }

or in .Net 3.5 and later:

    public void ExceedsThreshold(int threshold, 
            Action action, IEnumerable<bool> bools)
    { if (ExceedsThreshold(threshold, bools)) action(); }

or as an extension to IEnumerable<bool>

  public static class IEnumerableExtensions
  {
      public static bool ExceedsThreshold<T> 
         (this IEnumerable<bool> bools, int threshold)
      { return bools.Count(b => b) > threshold; }
  }

usage would then be:

  var bools = new [] {true, true, false, false, false, false, true};
  if (bools.ExceedsThreshold(3))
      // code to execute  ...

回答2:

I was going to write the Linq version, but five or so people beat me to it. But I really like the params approach to avoid having to manually new up an array. So I think the best hybrid is, based on rp's answer with the body replace with the obvious Linqness:

public static int Truth(params bool[] booleans)
{
    return booleans.Count(b => b);
}

Beautifully clear to read, and to use:

if (Truth(m, n, o, p, q) > 2)

回答3:

It's time for the obligatory LINQ answer, which in this case is actually quite neat.

var bools = new[] { true, true, false, false, false };

return bools.Count(b => b == true) > 1;

回答4:

I would just cast them to ints and sum.

Unless you're in a super tight inner loop, that has the benefit of being easy to understand.

回答5:

I'd write a function to receive any number of boolean values. It would return the number of those values that are true. Check the result for the number of values you need to be positive to do something.

Work harder to make it clear, not clever!

private int CountTrues( params bool[] booleans )
{
    int result = 0;
    foreach ( bool b in booleans )
    {
        if ( b ) result++;
    }

    return result;
}

回答6:

If there were millions instead of just 5 you could avoid Count()and do this instead ...

public static bool MoreThanOne (IEnumerable<bool> booleans)
{
    return booleans.SkipWhile(b => !b).Skip(1).Any(b => b);
}

回答7:

If your flags are packed into one word then Michael Burr's solution will work. However, the loop is not necessary:

int moreThanOneBitSet( unsigned int v)
{
    return (v & (v - 1)) != 0;
}

example

 v (binary) | v - 1 | v&(v-1) | result
------------+-------+---------+--------
       0000 |  1111 |    0000 |  false
       0001 |  0000 |    0000 |  false
       0010 |  0001 |    0000 |  false
       0011 |  0010 |    0010 |   true
       .... |  .... |    .... |   ....
       1000 |  0111 |    0000 |  false
       1001 |  1000 |    1000 |   true
       1010 |  1001 |    1000 |   true
       1011 |  1010 |    1010 |   true
       1100 |  1011 |    1000 |   true
       1101 |  1100 |    1100 |   true
       1110 |  1101 |    1100 |   true
       1111 |  1110 |    1110 |   true

回答8:

Shorter and uglier than Vilx-s version:

if (((a||b||c)&&(d||e))||((a||d)&&(b||c||e))||(b&&c)) {}

回答9:

if you mean more than or equal to one boolean equals to true, you could do it like

if (bool1 || bool2 || bool3 || bool4 || bool5)

If you need more than one (2 and above) booleans equal to true, you can try

int counter = 0;
if (bool1) counter++;
if (bool2) counter++;
if (bool3) counter++;
if (bool4) counter++;
if (bool5) counter++;
if (counter >= 2) //More than 1 boolean is true

回答10:

from the top of my head, a quick approach for this specific example; you could convert the bool to an int (0 or 1). then loop through therm and add them up. if the result >= 2 then you can execute your function.

回答11:

While I like LINQ, there are some holes in it, like this problem.

Doing a count is fine in general, but can become an issue when the items your counting take a while to calculate/retrieve.

The Any() extension method is fine if you just want to check for any, but if you want to check for at least there's no built in function that will do it and be lazy.

In the end, I wrote a function to return true if there are at least a certain number of items in the list.

public static bool AtLeast<T>(this IEnumerable<T> source, int number)
{
    if (source == null)
        throw new ArgumentNullException("source");

    int count = 0;
    using (IEnumerator<T> data = source.GetEnumerator())
        while (count < number && data.MoveNext())
        {
            count++;
        }
    return count == number;
}

To use:

var query = bools.Where(b => b).AtLeast(2);

This has the benefit of not needing to evaluate all the items before returning a result.

[Plug] My project, NExtension contains AtLeast, AtMost and overrides that allow you to mix in the predicate with the AtLeast/Most check. [/Plug]

回答12:

Casting to ints and summing should work, but it's a bit ugly and in some languages may not be possible.

How about something like

int count = (bool1? 1:0) + (bool2? 1:0) + (bool3? 1:0) + (bool4? 1:0) + (bool5? 1:0);

Or if you don't care about space, you could just precompute the truth table and use the bools as indices:

if (morethanone[bool1][bool2][bool3][bool4][bool5]) {
 ... do something ...
}

回答13:

I would do something like this, using the params argument.

        public void YourFunction()
        {
            if(AtLeast2AreTrue(b1, b2, b3, b4, b5))
            {
                // do stuff
            }
        }

        private bool AtLeast2AreTrue(params bool[] values)
        {
            int trueCount = 0;
            for(int index = 0; index < values.Length || trueCount >= 2; index++)
            {
                if(values[index])
                    trueCount++;
            }

            return trueCount > 2;

        }

回答14:

if (NumberOfTrue(new List<bool> { bool1, bool2, bool3, bool4 }) >= 2)
{
    // do stuff
}

int NumberOfTrue(IEnumerable<bool> bools)
{
    return bools.Count(b => b);
}

回答15:

Not exactly pretty... but here's another way to do it:

if (
    (a && (b || c || d || e)) ||
    (b && (c || d || e)) ||
    (c && (d || e)) ||
    (d && e)
)

回答16:

I have a much much better one now and very short!

bool[] bools = { b1, b2, b3, b4, b5 };
if (bools.Where(x => x).Count() > 1)
{
   //do stuff
}

回答17:

I wanted to give a C++11 variadic template answer.

template< typename T>
T countBool(T v)
{
    return v;
}

template< typename T, typename... Args>
int countBool(T first, Args... args)
{
    int boolCount = 0;
    if ( first )
        boolCount++;
    boolCount += countBool( args... );
    return boolCount;
}

simply calling it as follows creates a rather elegant method of counting the number of bools.

if ( countBool( bool1, bool2, bool3 ) > 1 )
{
  ....
}

回答18:

In most languages true is equivalent to a non-zero value while false is zero. I don't have exact syntax for you, but in pseudo code, what about:

if ((bool1 * 1) + (bool2 * 1) + (bool3 * 1) > 2)
{
    //statements here
}

回答19:

If you only have five different values, you can easily do the test by packing the bits in to a short or an int and checking to see if it is any of the zero or one bit answers. The only invalid numbers you could get would be..

0x 0000 0000 
0x 0000 0001
0x 0000 0010
0x 0000 0100
0x 0000 1000
0x 0001 0000

This gives you six values to search for, put them in a lookup table and if it's not in there, you have your answer.

This gives you a simple answer.

   public static boolean moreThan1BitSet(int b)
   {
      final short multiBitLookup[] = { 
            1, 1, 1, 0, 1, 0, 0, 0,
            1, 0, 0, 0, 0, 0, 0, 0,
            0, 0, 0, 0, 0, 0, 0, 0,
            1, 0, 0, 0, 0, 0, 0, 0,
            0, 0, 0, 0, 0, 0, 0, 0
      };
      if(multiBitLookup[b] == 1)
         return false;
      return true;
   }

This doesn't scale well past 8 bits, but you only have five.

回答20:

if((b1.CompareTo( false ) + b2.CompareTo( false ) + b3.CompareTo( false ) + ...) > 1)

// More than one of them are true

...

else

...

回答21:

You mentioned

One interesting option is to store the booleans in a byte, do a right shift and compare with the original byte. Something like if (myByte && (myByte >> 1))

I don't think that expression will give you the result you want (at least using C semantics, since the expression is not valid C#):

If (myByte == 0x08), then the expression will return true even though there's only one bit set.

If you meant "if (myByte & (myByte >> 1))" then if (myByte == 0x0a) the expression will return false even though there are 2 bits set.

But here are some techniques for counting the number of bits in a word:

Bit Twiddling Hacks - Counting bits

A variation you might consider is to use Kernighan's counting method, but bail out early since you only need to know if there's more than one bit set:

int moreThanOneBitSet( unsigned int v)
{
    unsigned int c; // c accumulates the total bits set in v

    for (c = 0; v && (c <= 1); c++)
    {
      v &= v - 1; // clear the least significant bit set
    }

    return (c > 1);
}

Of course, using a lookup table's not a bad option either.

回答22:

I was recently having this same issue, where I had three boolean values, which I needed to check that only 1 of them was true at a time. For this I used the xor operator as follows:

bool a = true;
bool b = true;
bool c = false;

if (a || b || c)
{
    if (a ^ b ^ c){
        //Throw Error
    }
}

This code will throw an error as a and b are both true.

For reference: http://www.dotnetperls.com/xor

I have only just found the xor operator in C# if anyone knows of any pit falls of this strategy, please let me know.