当引用相同的变量时,Swift Array内存地址会发生变化

我正在学习Python,并希望确认Objective-C和Swift中的某种行为.

测试如下:

蟒蛇

def replace(list):
    list[0] = 3
    print(list)

aList = [1, 2, 3]
print(aList)
replace(aList)
print(aList)

Objective-C的

- (void)replace:(NSMutableArray *)array {
    array[0] = @1;
    NSLog(@"array: %@, address: %p\n%lx", array, array, (long)&array);
}

NSMutableArray *array = [@[@1, @2, @3] mutableCopy];
NSLog(@"original: %@, address: %p \n%lx", array, array, (long)&array);
[self replace:array];
NSLog(@"modified: %@, address: %p \n%lx", array, array, (long)&array);

迅速

var numbers = [1, 2, 3]
let replace = { (var array:[Int]) -> Void in
    array[0] = 2
    print("array: \(array) address:\(unsafeAddressOf(array as! AnyObject))")
}

print("original: \(numbers) address:\(unsafeAddressOf(numbers as! AnyObject))")
replace(numbers)
print("modified: \(numbers) address:\(unsafeAddressOf(numbers as! AnyObject))")

除了Swift中的地址部分之外,所有结果都按预期发布.在Objective-C中,数组的地址在original和中保持不变modified,但Swift的打印结果是:

original: [1, 2, 3] address:0x00007f8ce1e092c0
array: [2, 2, 3] address:0x00007f8ce1f0c5d0
modified: [1, 2, 3] address:0x00007f8ce4800a10

有什么我想念的吗？

Swift中的数组具有值语义,而不是Python和Objective-C中数组的引用语义.您看到不同地址(以及地址)的原因是,每次进行as! AnyObject强制转换时,您实际上都在告诉Swift将您的Array结构桥接到一个实例NSArray.由于您桥接了三次,因此您将获得三个不同的地址.

您不应该考虑Swift数组的地址,但如果您想(暂时)获取数组缓冲区的地址,您可以这样做:

func getBufferAddress(array: [T]) -> String {
    return array.withUnsafeBufferPointer { buffer in
        return String(buffer.baseAddress)
    }
}

这使您可以看到缓冲区的写时复制:

var numbers = [1, 2, 3]
let numbersCopy = numbers

// the two arrays share a buffer here
getBufferAddress(numbers)                 // "0x00007fba6ad16770"
getBufferAddress(numbersCopy)             // "0x00007fba6ad16770"

// mutating `numbers` causes a copy of its contents to a new buffer
numbers[0] = 4

// now `numbers` has a new buffer address, while `numbersCopy` is unaffected
getBufferAddress(numbers)                 // "0x00007ff23a52cc30"
getBufferAddress(numbersCopy)             // "0x00007fba6ad16770"