Convert bit array to short

Question

I have a function that returns the bits of a short (inspired from Converting integer to a bit representation):

bool* bitsToShort(short value) {
    bool* bits = new bool[15];
    int count = 0;
    while(value) {
        if (value&1)
            bits[count] = 1;
        else
            bits[count] = 0;
        value>>=1;
        count++;
    }
    return bits;
}

How can I do the reverse? Converte the array of bits in the short?

With some bit/bitwise operators. Not sure if there's a better way/stdfunction — keyser
– keyser, Commented Dec 6, 2013 at 22:37
An implementation with 15-bit shorts does not conform to either the C or the C++ language definition. And even if it's changed to 16, that's a bad assumption to make without checking for overflow. Of course, that has nothing to do with the question. <g> — Pete Becker
– Pete Becker, Commented Dec 6, 2013 at 22:37
@PeteBecker An implementation where short has 15 value bits plus one sign bit does conform, and the code already doesn't deal with negative values properly (perhaps it doesn't need to) for other reasons, so if you ignore the sign bit, 15 bits remain. — user743382
– user743382, Commented Dec 6, 2013 at 22:45
@hvd - good point. The assumptions, then, are that value is non-negative and less than 2^16. — Pete Becker
– Pete Becker, Commented Dec 6, 2013 at 22:46
@PeteBecker Right, I can see that my comment wasn't very clear, but you got what I meant. — user743382
– user743382, Commented Dec 6, 2013 at 22:51

Ivan Smirnov · Accepted Answer · 2013-12-06 22:39:01Z

2

short shortFromBits(bool* bits) {
    short res = 0;
    for (int i = 0; i < 15; ++i) {
        if (bits[i]) {
            res |= 1 << i;
        }
    }
    return res;
}

res |= (1<<i) sets the i-th bit in res to one.

answered Dec 6, 2013 at 22:39

Ivan Smirnov

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2 Comments

user3055473 Over a year ago

Thanks I'll accept the answer, but just one more question: why is not 40000 work? give me back 7232.

mirabilos Over a year ago

Because 40000 is larger than a short. (40000 & 0x7FFF) == 7232.

Sergey Kalinichenko · Accepted Answer · 2013-12-07 00:02:33Z

1

Like this:

bool* bits = ... // some bits here
short res = 0;
for (int i = 14 ; i >= 0 ; i--) {
    res <<= 1;             // Shift left unconditionally
    if (bits[i]) res |= 1; // OR in a 1 into LSB when bits[i] is set
}
return res;

edited Dec 7, 2013 at 0:02

answered Dec 6, 2013 at 22:40

Sergey Kalinichenko

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1 Comment

mirabilos Over a year ago

In the code given, bits[0] is the LSB; in your code, bits[0] would be the MSB. Downvoted (sorry about that).

Pete Becker · Accepted Answer · 2013-12-06 22:50:01Z

-1

Essentially:

unsigned short value = 0;
for (int i = sizeof(unsigned short) * CHAR_BIT - 1; 0 <= i; --i) {
    value *= 2;
    if (bits[i)
        ++value;
}

This assumes that bits points to an array of bool with at least sizeof(unsigned short) elements. I have not tested it. There could be an off-by-one error somewhere. But maybe not.

edited Dec 6, 2013 at 22:50

answered Dec 6, 2013 at 22:42

Pete Becker

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