Swift: 如何确保代码没有被优化掉？

Question

我想将 Swift 中 UnsafeMutablePointer 的内容清零。

在 C 中你通常有这样的东西：

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);
}

如何在 Swift 中实现同样的目标？

// 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

好的，我在 Apple 开发者论坛上问了同样的问题，那里的一些人向我指出了 memset_s-函数，它可以满足我的要求。

所以我的 Swift-代码应该是这样的：

// 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)
}

Swift: 如何确保代码没有被优化掉？

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

volatile

compiler-optimization

swift