That Crazy Carl Guy - 10 months ago 88

C++ Question

I am trying to generate a hash code for an object in 3D space so it can be quickly found in an array using a binary search algorithm.

Since each object in this array has a unique XYZ location, I figured I could use those three values to generate the hash code. I used the following code to try and generate the hash code.

`int64_t generateCode(int16_t x, int16_t y, int16_t z) {`

int64_t hashCode = z;//Set Z bits.

hashCode <<= 16;//Shift them 16 bits.

hashCode |= y;//Set Y bits.

hashCode <<= 16;//Shift them 16 bits.

hashCode |= x;//Set X bits.

}

Now here is the problem from what I can tell. Consider the following peace of code:

`int16_t x = -1;`

cout << "X: " << bitset<16>(x) << endl;//Prints the binary value of X.

int64_t y = x;//Set Y to X. This will automatically cast the types.

cout << "Y: " << bitset<64>(y) << endl;//Prints the binary value of Y.

The output of this program is as follows:

`X: 1111111111111111`

Y: 1111111111111111111111111111111111111111111111111111111111111111

It keeps the numerical value of the number, but changes the underlying binary to do that. I don't want to modify that binary so I can have an output like the following:

`X: 1111111111111111`

Y: 0000000000000000000000000000000000000000000000001111111111111111

By doing that, I can then create a unique hash code from the XYZ values that would look like the following:

`Unused X Y Z`

HashCode: [0000000000000000][0000000000000000][0000000000000000][0000000000000000]

And that will be used for the binary search.

Answer Source

Convert the `int16_t`

to a `uint16_t`

first, then merge them together into a `uint64_t`

that you finally cast to a `int64_t`

:

```
int64_t generateCode(int16_t x, int16_t y, int16_t z) {
uint64_t hashCode = static_cast<uint16_t>(z);
hashCode <<= 16;
hashCode |= static_cast<uint16_t>(y);
hashCode <<= 16;
hashCode |= static_cast<uint16_t>(x);
return static_cast<int64_t>(hashCode);
}
```

The `int16_t`

/`int64_t`

types will be a two's complement representation (7.20.1.1 paragraph 1 of the C standard requires this), so converting them to a `uint*_t`

of the same size will be a bit-wise no-op.