Converting a C char array to a String

Question

I have a Swift program that does interop with a C library. This C library returns a structure with a char[] array inside, like this:

struct record
{
    char name[8];
};

The definition is correctly imported into Swift. However, the field is interpreted as a tuple of 8 Int8 elements (typed (Int8, Int8, Int8, Int8, Int8, Int8, Int8, Int8)), which I have no idea how to transform into a String with Swift.

There is no String initializer that accepts an Int8 tuple, and it doesn't seem possible to get a pointer to the first element of the tuple (since types can be heterogenous, that's not really surprising).

Right now, my best idea is to create a tiny C function that accepts a pointer to the structure itself and return name as a char* pointer instead of an array, and go with that.

Is there, however, are pure Swift way to do it?

Answer 1

The C array char name[8] is imported to Swift as a tuple:

(Int8, Int8, Int8, Int8, Int8, Int8, Int8, Int8)

The address of name is the same as the address of name[0], and Swift preserves the memory layout of structures imported from C, as confirmed by Apple engineer Joe Groff:

... You can leave the struct defined in C and import it into Swift. Swift will respect C's layout.

As a consequence, we can pass the address of record.name, converted to an UInt8 pointer, to the String initializer. The following code has been updated for Swift 4.2 and later:

let record = someFunctionReturningAStructRecord()
let name = withUnsafePointer(to: record.name) {
    $0.withMemoryRebound(to: UInt8.self, capacity: MemoryLayout.size(ofValue: $0)) {
        String(cString: $0)
    }
}

NOTE: It is assumed that the bytes in name[] are a valid NUL-terminated UTF-8 sequence.

For older versions of Swift:

// Swift 2:
var record = someFunctionReturningAStructRecord()
let name = withUnsafePointer(&record.name) {
    String.fromCString(UnsafePointer($0))!
}

// Swift 3:
var record = someFunctionReturningAStructRecord()
let name = withUnsafePointer(to: &record.name) {
    $0.withMemoryRebound(to: UInt8.self, capacity: MemoryLayout.size(ofValue: record.name)) {
        String(cString: $0)
    }
}

Answer 2

You can actually collect a tuple into an array by using Swift's variadic parameter syntax:

let record = getRecord()
let (int8s: Int8...) = myRecord          // int8s is an [Int8]
let uint8s = int8s.map { UInt8($0) }
let string = String(bytes: uint8s, encoding: NSASCIIStringEncoding)
// myString == Optional("12345678")

Answer 3

I'm interested in working this out for my own purposes as well, so I added a new function:

func asciiCArrayToSwiftString(cString:Int8...) -> String
{
    var swiftString = String()            // The Swift String to be Returned is Intialized to an Empty String
    var workingCharacter:UnicodeScalar = UnicodeScalar(UInt8(cString[0]))
    var count:Int = cString.count

    for var i:Int = 0; i < count; i++
    {
        workingCharacter = UnicodeScalar(UInt8(cString[i])) // Convert the Int8 Character to a Unicode Scalar
        swiftString.append(workingCharacter)             // Append the Unicode Scalar

    }

    return swiftString                     // Return the Swift String
}

I call this function with:

    let t:Int8 = Int8(116)
    let e:Int8 = Int8(101)
    let s:Int8 = Int8(115)
    let testCString = (t, e, s, t)
    let testSwiftString = wispStringConverter.asciiCArrayToSwiftString(testCString.0, testCString.1, testCString.2, testCString.3)
    println("testSwiftString = \(testSwiftString)")

the resulting output is:

testSwiftString = test

Answer 4

I have just experienced a similar issue using Swift 3. (3.0.2). I was attempting to convert an Array of CChar, [CChar] to a String in Swift. It turns out Swift 3 has a String initializer which will take a cString.

Example:

let a = "abc".cString(using: .utf8) // type of a is [CChar]
let b = String(cString: a!, encoding: .utf8) // type of b is String
print("a = \(a)")
print("b = \(b)")

results in

a = Optional([97, 98, 99, 0])

b = Optional("abc")

Note that the cString function on String results in an Optional. It must be force unwrapped when used in the String.init function creating b. And b is also Optional... meaning both could end up being nil, so error checking should also be used.