C++ Bitset << operator not working. Pointer to bitset variable

Question

I have a set of bitsets pointers in an unordered_map

static unordered_map< size_t, bitset<BITSIZE>* > systemBits;

And my function

template<typename system>
     static bitset<BITSIZE> & getBitFor() {

    size_t hash = typeid(system).hash_code();

    bitset<BITSIZE> * bit = systemBits[hash];

    if(bit == NULL) {
        bit = new bitset<BITSIZE>(0);
        (*bit) << POS++; // tried *bit << POS++ as well;
        systemBits[hash] = bit;
    }

    return *bit;
}

Whereas POS is an int set to 1 at first.

All that I'm trying to do is shift the bitset with the amounts of position per new bitset.

 (*bit) << POS++;

However this doesn't seem to work. When I cout the returned bitset all its bits are set to 0. If I do to the following:

bit->flip();

or (*bit).flip();

The bitset returned does flip all the 0 to 1.

What gives? Why does the shift operator have no effect at all?

Even so, I'm calling the function twice on two different object. So it should at least print one bitset with 1 shift. But no. POS = 1; — Sidar, Jun 26 '12 at 08:46

Oliver Charlesworth · Accepted Answer · 2012-06-26T08:52:08.833

0

Because you're not assigning the result to anything . You'd have the same problem if you were trying to shift an int.

Also, you've initialized your bitset to zero, and zero shifted by any amount is always zero.

edited Jun 26 '12 at 08:52

answered Jun 26 '12 at 08:43

Oliver Charlesworth

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Why does it need assignment? I'm working with the bitset directly. – Sidar Jun 26 '12 at 08:45
@sidar for the same reason that int i= 3; i<< 2; doesn't do anything. – Oliver Charlesworth Jun 26 '12 at 08:49
Well that was majorly lame. (*bit) = 1 << POS++; did the trick. Now I'm limited to the bit size of a number type though. Need to use a nother bitset then. Thanks mate. – Sidar Jun 26 '12 at 08:51

C++ Bitset << operator not working. Pointer to bitset variable

1 Answers1