Swift: How to make sure that code is not optimized out? - swift

Question

I want to zero-out the contents of an UnsafeMutablePointer in Swift.
In C you usually have something like this:
void freeSecure(void *buffer, uint64_t size) {
// Create volatile pointer to make sure that the code won't be optimized away
volatile uint8_t *ptr = buffer;
for (uint64_t i = 0; i < size; i++) ptr[i] = 0x00;
free(buffer);
}
How can I achieve the same thing in Swift?
// Is extension of `UnsafeMutablePointer`
public func KC_dealloc(allocated: Int) {
if num == 0 {
self.destroy()
self.dealloc(allocated)
return
}
let byteCount = sizeof(Memory) * allocated
let ptr = UnsafeMutablePointer<UInt8>(self) // volatile???
for var i = 0; i < byteCount; i++ {
ptr[i] = 0x00
}
self.destroy()
self.dealloc(allocated)
}

Answer 1

Okay, I asked the same question in the Apple-developer-forum and some guy there pointed me to the memset_s-function which does what I want.
So my Swift-code should look like this:
// Is extension of `UnsafeMutablePointer`
public func KC_dealloc(allocated: Int) {
if num == 0 {
self.destroy()
self.dealloc(allocated)
return
}
let byteCount = sizeof(Memory) * allocated
let ptr = UnsafeMutablePointer<UInt8>(self) // volatile???
memset_s(ptr, byteCount, 0x00, byteCount) // Defined in C11
self.destroy()
self.dealloc(allocated)
}