Bitwise concatenation of two numbers

Question

I intend to write a function which accepts two numbers and concatenates their binary values such that:

resultingNumber=(number1)|(number2)

Here i need ((number1) << num_bits_in_2) |(number2). i.e. number2 should occupy the lowest bits. Number1 should occupy the higher bits and should not overlap with bits of maxnumber2

func(number1, number2, maxnumber1, maxnumber2)
{
    long rem,i=1,binaryMaxNumber1=0,binaryMaxNumber2=0, lenNumber1=0, lenNumber2=0;
    do
    {
        rem=maxNumber1%2;
        binaryMaxNumber1=binaryMaxNumber1 + (i*rem);
        lenNumber1++;
        maxNumber1=maxNumber1/2;
        i=i*10;
    }while(maxNumber1>0);

    i=1;
    do
    {
        rem=maxNumber2%2;
        binaryMaxNumber2=binaryMaxNumber2 + (i*rem);
        lenNumber2++;
        maxNumber2=maxNumber2/2;
        i=i*10;
    }while(maxNumber2>0);

    unsigned resultingNumber=(number1<<lenNumber2)|(number2);
}

I wrote the above mentioned program in order to get the desired results. Is there any other way which is more efficient than this to achieve the same in C/c++. As I need to call this function again and again for very large numbers.

Is this a C question or a C++ question? There is no such language as "C/C++". — Fred Larson, Jul 31 '14 at 15:15
is `lenNumber` just going to be equal to position of `maxNumber`'s leftmost set bit? You don't even use `rem` and `binaryMaxNumber` either — Ben, Jul 31 '14 at 15:15
`((number2) << num_bits_in_1) |(number1)` or am I missing something? — Neil Kirk, Jul 31 '14 at 15:16
@NeilKirk Its vice versa: ((number1) << num_bits_in_2) |(number2). i.e. number2 should occupy the lowest bits. Number1 should occupy the highest bits but should not overlap with bits of maxnumber2 — user3809749, Jul 31 '14 at 15:20
@Ben Yes I dont use it..therefore the code is inefficient but i need them in order to find the number of bits by which I should perform bit shift — user3809749, Jul 31 '14 at 15:22
@user3809749 Your own comment appears to mostly satisfy the post `((number1) << num_bits_in_2) |(number2)`. What else do you need? — chux - Reinstate Monica, Jul 31 '14 at 15:26
possible duplicate of [How to write log base(2) in c/c++](http://stackoverflow.com/questions/3064926/how-to-write-log-base2-in-c-c) — anatolyg, Jul 31 '14 at 15:36

score 3 · Accepted Answer · answered Jul 31 '14 at 15:33

3

Assuming your codes is correct your function can be rewritten like this:

unsigned long func( unsigned long number1, unsigned long number2, unsigned long , unsigned long maxnumber2)
{
    while( maxnumber2 ) {
        maxnumber2 >>= 1;
        number1 <<= 1;
    }
    return number1 | number2;
}

But if you use the same maxnumber2 it would be better to count number of bits to shift outside of the function and pass that instead of maxnumber.

answered Jul 31 '14 at 15:33

Slava

43,454
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Better yet, if `maxnumber2` is constant, use a `static const int bitsnumber2`. – Khouri Giordano Jul 31 '14 at 15:39

score 1 · Answer 2 · answered Jul 31 '14 at 15:42

1

You can use _BitScanReverse intrinsic to quickly find the number of bits for number2.

long long func( long number1, long number2)
{
    unsigned long numBits = 0;
    if( _BitScanReverse(&numBits, (unsigned)number1))
    {
        return (long long)(number1<<numBits)|number2;
    }
    return (long long)number2;
}

Note that the size of the resulting number should be big enough to contain the other numbers.

answered Jul 31 '14 at 15:42

rashmatash

1,699
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number of bits in `maxnumber2` not `number2`. And where did OP say that this is (only) for MS VS? – Slava Jul 31 '14 at 15:44

Bitwise concatenation of two numbers

2 Answers2