How to operate on LLVM Bitvectors?

Question

(I apologize if this is straight forward, I'm not well versed in C++)

I have

#include <llvm/ADT/BitVector.h>

Looking at the defintion file, it defines

// Intersection, union, disjoint union
bitVector &operator&=(const BitVector RHS) { ... }

I want to do the union of two bit vectors; that seems to be a definition for intersection, so I'm guessing intersection would be something like this:

bitVector1 &= bitVector2;

Is this correct? Further down, it declares, without comment,

bitVector &operator|=(const bitVector &RHS) { ... }

Would it correct to assume this is the union operator? (bit operation is |=)

Thanks for the help!

Answer 1

The &= performs logical AND on the individual elements of the LHS and RHS BitVector operands and stores the result in LHS operand.

The |= performs logical OR on the individual elements of the LHS and RHS BitVector operands and stores the result in LHS operand.

A minimal example:

#include <iostream>
#include <llvm/ADT/BitVector.h>

llvm::BitVector  bitVector1(10, true);
llvm::BitVector  bitVector2(10, false);
llvm::BitVector  bitVector3(10, true);

int main()
{
    for(int i = 0; i < bitVector1.size(); i++)
        std::cout << bitVector1[i] << ' ';
    std::cout << '\n';        

    bitVector2[4] = true; //5th element set to true

    bitVector1 &= bitVector2;

    for(int i = 0; i < bitVector1.size(); i++)
        std::cout << bitVector1[i] << ' ';
    std::cout << '\n';


    bitVector1 |= bitVector3;


    for(int i = 0; i < bitVector1.size(); i++)
        std::cout << bitVector1[i] << ' ';
    std::cout << '\n';        

}

Output:

1 1 1 1 1 1 1 1 1 1 
0 0 0 0 1 0 0 0 0 0 
1 1 1 1 1 1 1 1 1 1 

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