All combinations of 0's and 1's in 128 bit character array?

Quote:

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Originally Posted by itsmeandnobodyelse

The vector already holds an internal 16-byte elements array.

If you do

Code:

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    unsigned char * p = (unsigned char *)&v[0];
    some_otherfunction_expecting_a_16bytes_array(p);

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where v is one of the permutations you got the pointer p would point to a 16-elements array.

If you want to copy the array to fixed-sized char array, you would/could do

Code:

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    unsigned char arr[16] = { '\0' };
    unsigned char * p = (unsigned char *)&v[0];
    memcpy(arr, p, min(sizeof(arr), v.size() * sizeof(v[0])));

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The min is only to make sure that the vector really has 16 elements of type char, same as the arr.

Regards, Alex

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That is all very true, but the problem remains that if you store bytes inside the vector, when you call next_permutation, you will be per mutating bytes, and not bits. Sure, you can recover the chars easily, but they won't be properlly shifted

A good solution would have been to use a vector<bool>, as it stores bit(-like) objects. This would work perfectly if you could write &v[0] on a vector bool, but you can't (at least not to my knowledge).

IMO, it's a lose-lose situation, where you have to get creative and daisy chain several solutions.

Quote:

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Originally Posted by monarch_dodra

That is all very true, but the problem remains that if you store bytes inside the vector, when you call next_permutation, you will be per mutating bytes, and not bits.

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So what? Make your container 8x larger, and you'll have logically the same result.

Quote:

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Originally Posted by Lindley

So what? Make your container 8x larger, and you'll have logically the same result.

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Yes, if you make your container 8 times larger, you get the desired result, but from there, it's the copying back into the char array that becomes problematic :/

Quote:

---

Originally Posted by itsmeandnobodyelse

The vector already holds an internal 16-byte elements array.

If you do

Code:

---

    unsigned char * p = (unsigned char *)&v[0];
    some_otherfunction_expecting_a_16bytes_array(p);

---

where v is one of the permutations you got the pointer p would point to a 16-elements array.

If you want to copy the array to fixed-sized char array, you would/could do

Code:

---

    unsigned char arr[16] = { '\0' };
    unsigned char * p = (unsigned char *)&v[0];
    memcpy(arr, p, min(sizeof(arr), v.size() * sizeof(v[0])));

---

The min is only to make sure that the vector really has 16 elements of type char, same as the arr.

Regards, Alex

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:( the index should be 15 :cry:

Hmm, actually, using bitset instead of a vector and then just memcpying into a char array of length 16 (bytes) works :)

Thanks all!

Yikes. You shouldn't use memcpy() on C++ objects.

If it's just an assignment, whatever. But if you actually intend this code to be portable and robust, that's not a good approach.