C++ – std::set::insert, duplicate elements

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What would be an efficient implementation for a std::set insert member function? Because the data structure sorts elements based on std::less (operator < needs to be defined for the element type), it is conceptually easy to detect a duplicate.

How does it actually work internally? Does it make use of the red-back tree data structure (a mentioned implementation detail in the book of Josuttis)?

Implementations of the standard data structures may vary…

I have a problem where I am forced to have a (generally speaking) sets of integers which should be unique. The length of the sets varies so I am in need of dynamical data structure (based on my narrow knowledge, this narrows things down to list, set). The elements do not necessarily need to be sorted, but there may be no duplicates. Since the candidate sets always have a lot of duplicates (sets are small, up to 64 elements), will trying to insert duplicates into std::set with the insert member function cause a lot of overhead compared to std::list and another algorithm that may not resort to having the elements sorted?

Additional: the output set has a fixed size of 27 elements. Sorry, I forgot this… this works for a special case of the problem. For other cases, the length is arbitrary (lower than the input set).

Best Answer

If you're creating the entire set all at once, you could try using std::vector to hold the elements, std::sort to sort them, and std::unique to prune out the duplicates.

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – Concatenating two std::vectors

vector1.insert( vector1.end(), vector2.begin(), vector2.end() );