是否有任何方式来模拟[NSString stringWithFormat:@“%p”,myVar],从Objective-C,在新的Swift语言?
例如:
let str = "A String"
println(" str value \(str) has address: ?")
是否有任何方式来模拟[NSString stringWithFormat:@“%p”,myVar],从Objective-C,在新的Swift语言?
例如:
let str = "A String"
println(" str value \(str) has address: ?")
当前回答
在Swift4 about Array中:
let array1 = [1,2,3]
let array2 = array1
array1.withUnsafeBufferPointer { (point) in
print(point) // UnsafeBufferPointer(start: 0x00006000004681e0, count: 3)
}
array2.withUnsafeBufferPointer { (point) in
print(point) // UnsafeBufferPointer(start: 0x00006000004681e0, count: 3)
}
其他回答
在Swift4 about Array中:
let array1 = [1,2,3]
let array2 = array1
array1.withUnsafeBufferPointer { (point) in
print(point) // UnsafeBufferPointer(start: 0x00006000004681e0, count: 3)
}
array2.withUnsafeBufferPointer { (point) in
print(point) // UnsafeBufferPointer(start: 0x00006000004681e0, count: 3)
}
其他答案都很好,尽管我正在寻找一种方法来获取整数形式的指针地址:
let ptr = unsafeAddressOf(obj)
let nullPtr = UnsafePointer<Void>(bitPattern: 0)
/// This gets the address of pointer
let address = nullPtr.distanceTo(ptr) // This is Int
只是跟进一下。
我对Swift 3的解决方案
extension MyClass: CustomStringConvertible {
var description: String {
return "<\(type(of: self)): 0x\(String(unsafeBitCast(self, to: Int.self), radix: 16, uppercase: false))>"
}
}
这段代码创建的描述类似于默认描述 < MyClass: 0 x610000223340 >
如果你只是想在调试器中看到这个,而不做任何其他事情,实际上没有必要获得Int指针。要在内存中获取对象地址的字符串表示形式,只需使用如下方法:
public extension NSObject { // Extension syntax is cleaner for my use. If your needs stem outside NSObject, you may change the extension's target or place the logic in a global function
public var pointerString: String {
return String(format: "%p", self)
}
}
使用示例:
print(self.pointerString, "Doing something...")
// Prints like: 0x7fd190d0f270 Doing something...
此外,请记住,你可以简单地打印一个对象,而不重写它的描述,它将显示它的指针地址与更多的描述性(如果经常是隐晦的)文本。
print(self, "Doing something else...")
// Prints like: <MyModule.MyClass: 0x7fd190d0f270> Doing something else...
// Sometimes like: <_TtCC14__lldb_expr_668MyModule7MyClass: 0x7fd190d0f270> Doing something else...
就用这个吧:
print(String(format: "%p", object))