Convert String to UnsafeMutablePointer<char_t> in Swift

Question

I'm working with a third party c API I'm trying to call one of the functions with a simple string. Something like this:

some_c_func("aString");

I get a build error:

Type 'UnsafeMutablePointer<char_t>' does not conform to protocol 'StringLiteralConvertible'

I've seen some suggestions to use utf8 on String or similar conversions, which gets nearly there, but with the following error:

some_c_func("aString".cStringUsingEncoding(NSUTF8StringEncoding));
'UnsafePointer<Int8>' is not convertible to 'UnsafeMutablePointer<char_t>'

How can I create an UnsafeMutablePointer?

Answer 1

It all depends on what char_t is.

If char_t converts to Int8 then the following will work.

if let cString = str.cStringUsingEncoding(NSUTF8StringEncoding) {
    some_c_func(strdup(cString))
}

This can be collapsed to

some_c_func(strdup(str.cStringUsingEncoding(NSUTF8StringEncoding)!))

WARNING! This second method will cause a crash if func cStringUsingEncoding(_:) returns nil.

---

Updating for Swift 3, and to fix memory leak

If the C string is only needed in a local scope, then no strdup() is needed.

guard let cString = str.cString(using: .utf8) else {
    return
}

some_c_func(cString)

cString will have the same memory lifecycle as str (well similar at least).

If the C string needs to live outside the local scope, then you will need a copy. That copy will need to be freed.

guard let interimString = str.cString(using: .utf8), let cString = strdup(interimString) else {
    return
}

some_c_func(cString)

//…

free(cString)

Answer 2

it may be simpler than that - many C APIs pass strings around as char * types, and swift treats these as unsafe.

try updating the C API (good) or hack it's header files (bad) to declare these as const char * instead.

in my experience this allows you to pass standard swift String types directly to the C API.

apparently a constant is required, in order to conform to the protocol.