Swift中的withUnsafePointer和copy on write验证
在Swift编程中,我们经常需要使用withUnsafePointer函数来获取指向值的指针。下面是一个示例代码:
withUnsafeBufferPointer(to: a) { point in
let address = UnsafeRawPointer(point)
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
}
对于值类型的赋值,Swift中的容器类型(如Array、Set)采用了copy on write(写时复制)的机制进行优化。这意味着在赋值操作时,并不会立即进行拷贝,只有在修改值时才会发生拷贝。这种优化可以避免不必要的性能损失。
对于Array、Dictionary、Set类型,当它们赋值的时候不会发生拷贝,只有在修改的之后才会发生拷贝。
通过代码验证,我们可以发现在使用withUnsafePointer函数时,数组赋值后两个变量的地址会不一样,这是合理的,因为值类型的赋值会使得两个变量在内存上互为独立个体。如果要验证copy on write机制,Array提供了另一个方法,即withUnsafeBufferPointer。使用该方法查看赋值后的数组,会发现输出的地址不变。
实际上,withUnsafeBufferPointer指向的地址是数组存储数据的内存地址,而withUnsafePointer指向的是数组对象本身。
以下是一个代码示例:
var array1 = [1, 2, 3]
array1.withUnsafeBufferPointer { point in
let address = point.baseAddress!
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 105553141216992
}
withUnsafePointer(to: array1) { point in
let address = UnsafeRawPointer(point)
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 6089438336
}
var array2 = array1 // 这里并没有真正复制数据,而是共享内存
array2.withUnsafeBufferPointer { point in
let address = point.baseAddress!
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 105553141216992
}
withUnsafePointer(to: array2) { point in
let address = UnsafeRawPointer(point)
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 6089438320
}
array2[2] = 1
array2.withUnsafeBufferPointer { point in
let address = point.baseAddress!
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 105553141218080
}
withUnsafePointer(to: array2) { point in
let address = UnsafeRawPointer(point)
let addressInt = Int(bitPattern: address)
print("\(addressInt)")
// 6089438280
}
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