How to suppress a specific warning in Swift

Question

I have a Swift function doing something like this:

func f() -> Int {
    switch (__WORDSIZE) {
        case 32: return 1
        case 64: return 2
        default: return 0
    }
}

Because __WORDSIZE is a constant, the compiler always gives at least one warning in the switch body. Which lines are actually marked depends on the target I am building for (e.g. iPhone 5 vs. 6; interestingly iPhone 5 gives a warning for the 64-bit case whereas iPhone 6 gives two warnings for 32-bit and default).

I found out that the Swift equivalent for #pragma is // MARK:, so I tried

// MARK: clang diagnostic push
// MARK: clang diagnostic ignored "-Wall"
func f() -> Int {
    switch (__WORDSIZE) {
        case 32: return 1
        case 64: return 2
        default: return 0
    }
}
// MARK: clang diagnostic pop

but the warnings remain, the MARKs seem to have no effect.

As a workaround, I now have something like this:

#if arch(arm) || arch(i386)
    return 1
#else
    #if arch(arm64) || arch(x86_64)
        return 2
    #else
        return 0
    #endif
#endif

– but of course this is not the same. Any hints…?

Answer 1

At present (Xcode 7.1), there seems to be no way of suppressing a specific warning in Swift (see e.g. How to silence a warning in swift).

In your special case, you can fool the compiler by computing the number of bytes in a word:

func f() -> Int {
    switch (__WORDSIZE / CHAR_BIT) { // Or: switch (sizeof(Int.self))
    case 4: return 1
    case 8: return 2
    default: return 0
    }
}

This compiles without warnings on both 32-bit and 64-bit architectures.