It took me longer to format this to paste it in here than it did to generate it.

Convert the following code into Xojo

```
int32_t intVal=0x12345678;
uint16_t lowWord=(intVal & 0xFFFF);
uint16_t highWord=((intVal>>16)) & 0xFFFF);
```

```
Dim intVal As Int32 = &h12345678
Dim lowWord As UInt16 = intVal And &hFFFF
Dim highWord As UInt16 = (intVal >> 16) And &hFFFF
```

Xojo doesn’t use >> for shifting

You’re correct! Xojo uses the `Bitwise.ShiftRight`

operator to perform a bitwise right shift. Here’s the updated code:

```
Dim intVal As Int32 = &h12345678
Dim lowWord As UInt16 = intVal And &hFFFF
Dim highWord As UInt16 = Bitwise.ShiftRight(intVal, 16) And &hFFFF
```

This should give you the same result as the original C++ code.

Please explain the code

Certainly! Let me break down the code line by line:

```
Dim intVal As Int32 = &h12345678
```

This line declares a variable `intVal`

of type `Int32`

(a 32-bit integer), and assigns it the hexadecimal value `&h12345678`

.

```
Dim lowWord As UInt16 = intVal And &hFFFF
```

This line declares a variable `lowWord`

of type `UInt16`

(a 16-bit unsigned integer), and assigns it the result of a bitwise AND operation between `intVal`

and the hexadecimal value `&hFFFF`

(which is equivalent to binary `1111111111111111`

). This operation effectively extracts the lower 16 bits (the “low word”) of `intVal`

.

```
Dim highWord As UInt16 = Bitwise.ShiftRight(intVal, 16) And &hFFFF
```

This line declares a variable `highWord`

of type `UInt16`

, and assigns it the result of two operations. The first operation is a bitwise right shift of `intVal`

by 16 bits, which effectively moves the upper 16 bits (the “high word”) of `intVal`

into the lower 16 bits. The second operation is a bitwise AND with `&hFFFF`

, which effectively clears the upper 16 bits and leaves only the lower 16 bits. This gives us the upper 16 bits (the “high word”) of the original `intVal`

.

So, overall, these three lines of code extract the lower 16 bits and upper 16 bits of a 32-bit integer value, using bitwise operations to mask and shift the bits as needed.

Reverse the process to end up with an Int32

Sure thing! Here’s how you can reverse the process to combine the `lowWord`

and `highWord`

variables into an `Int32`

:

```
Dim lowWord As UInt16 = &h5678
Dim highWord As UInt16 = &h1234
Dim intVal As Int32 = Bitwise.ShiftLeft(highWord, 16) Or lowWord
```

This code assumes you have already extracted the lower 16 bits (`lowWord`

) and upper 16 bits (`highWord`

) of an `Int32`

value, and assigns them to variables of type `UInt16`

.

The first line declares a variable `lowWord`

and assigns it the value `&h5678`

(which is the hexadecimal representation of the lower 16 bits).

The second line declares a variable `highWord`

and assigns it the value `&h1234`

(which is the hexadecimal representation of the upper 16 bits).

The third line uses bitwise operations to combine `lowWord`

and `highWord`

into a single `Int32`

. The `Bitwise.ShiftLeft`

operator shifts the bits in `highWord`

16 positions to the left, effectively moving them to the upper 16 bits of the resulting `Int32`

. The `Or`

operator then combines the shifted `highWord`

with `lowWord`

, effectively combining the upper and lower 16 bits into a single 32-bit value.

So, overall, these three lines of code reverse the process of extracting the lower and upper 16 bits from an `Int32`

, and combine them back into a single `Int32`

value.