fun*_*ct7 3 pointers memory-address ios swift
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)
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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);
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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))")
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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
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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<Int>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<T>(array: [T]) -> String {
return array.withUnsafeBufferPointer { buffer in
return String(buffer.baseAddress)
}
}
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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"
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