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swiftbinarytype-conversionunsigned-integer

Binary operator '|' cannot be applied to operands of type 'Int' and 'UInt8'

I wish to perform simple or logic on 2 bit maps, yet Swift thinks this is wrong:

let u: UInt8 = 0b1
let i: Int = 0b10
i | u // Binary operator '|' cannot be applied to operands of type 'Int' and 'UInit8'

Any way to conform to type inference and still have this working?

I could always do i | Int(u) // 3 but this, I think, is not optimal.

Solution

  • One of the fundamental principles of Swift is that is does not implicitly convert between types.

    let u: UInt8 = 0b1
let i: Int = 0b10
i | Int(u)

    forces you to think about the necessary conversions and what type the result should have, so that is the correct solution (in my opinion).

    Let's consider another example:

    let u: UInt8 = 128
let i: Int8 = -128

    What should u|i be? Both

    u | UInt8(i)
Int8(u) | i

    crash at runtime because u is not in the range of an Int8, and i is not in the range of an UInt8. Both 

    u | UInt8(bitPattern: i)   // result is `UInt8`
Int8(bitPattern: u) | i    // result is `Int8`

    would work, but how should the compiler choose between both? One could convert both to some larger type, for example

    Int(u) | Int(i)

    but that type is somewhat arbitrary, how could it automatically be inferred from the compiler? And what is the "larger type" for Int64 + UInt64 arguments?

    That's why I think that an explicit conversion is the right solution.