Convert a two byte UInt8 array to a UInt16 in Swift

Question

With Swift I want to convert bytes from a uint8_t array to an integer.

\"C\" Example:

char bytes[2] = {0x01, 0x02};
NSData *data = [NSData dataWithBy         


        

Answer 1

Assuming little endian encoding.

To convert to UInt16 from [UInt8], you can do something like

var x: [UInt8] = [0x01, 0x02]
var y: UInt16 = 0
y += UInt16(x[1]) << 0o10
y += UInt16(x[0]) << 0o00

For conversion to UInt32, this pattern extends to

var x: [UInt8] = [0x01, 0x02, 0x03, 0x04]
var y: UInt32 = 0
y += UInt32(x[3]) << 0o30
y += UInt32(x[2]) << 0o20
y += UInt32(x[1]) << 0o10
y += UInt32(x[0]) << 0o00

Octal representation of the shift amount gives a nice indication on how many full bytes are shifted (8 becomes 0o10, 16 becomes 0o20 etc).

This can be reduced to the following for UInt16:

var x: [UInt8] = [0x01, 0x02]
let y: UInt16 = reverse(x).reduce(UInt16(0)) {
    $0 << 0o10 + UInt16($1)
}

and to the following for UInt32:

var x: [UInt8] = [0x01, 0x02, 0x03, 0x04]
let y: UInt32 = reverse(x).reduce(UInt32(0)) {
    $0 << 0o10 + UInt32($1)
}

The reduced version also works for UInt64, and also handles values where the byte encoding does not use all bytes, like [0x01, 0x02, 0x03]