equality

问题 Can someone please explain to me why the output from the following code is saying that arrays are not equal ? int main() { int iar1[] = {1,2,3,4,5}; int iar2[] = {1,2,3,4,5}; if (iar1 == iar2) cout << "Arrays are equal."; else cout << "Arrays are not equal."; return 0; } 回答1: if (iar1 == iar2) Here iar1 and iar2 are decaying to pointers to the first elements of the respective arrays. Since they are two distinct arrays, the pointer values are, of course, different and your comparison tests not

问题 Can someone please explain to me why the output from the following code is saying that arrays are not equal ? int main() { int iar1[] = {1,2,3,4,5}; int iar2[] = {1,2,3,4,5}; if (iar1 == iar2) cout << "Arrays are equal."; else cout << "Arrays are not equal."; return 0; } 回答1: if (iar1 == iar2) Here iar1 and iar2 are decaying to pointers to the first elements of the respective arrays. Since they are two distinct arrays, the pointer values are, of course, different and your comparison tests not

问题 I had a bug where I was relying on methods being equal to each other when using is . It turns out that's not the case: >>> class What(object): def meth(self): pass >>> What.meth is What.meth False >>> inst = What() >>> inst.meth is inst.meth False Why is that the case? It works for regular functions: >>> def func(): pass >>> func is func True 回答1: Method objects are created each time you access them . Functions act as descriptors, returning a method object when their .__get__ method is called

问题 I'm reading through Programming in Scala. It says: You can redefine the behavior of == for new types by overriding the equals method, which is always inherited from class Any . The inherited equals , which takes effect unless overridden, is object identity, as is the case in Java. So equals (and with it, == ) is by default the same as eq , but you can change its behavior by overriding the equals method in the classes you define. It is not possible to override == directly, as it is defined as

问题 I'm reading through Programming in Scala. It says: You can redefine the behavior of == for new types by overriding the equals method, which is always inherited from class Any . The inherited equals , which takes effect unless overridden, is object identity, as is the case in Java. So equals (and with it, == ) is by default the same as eq , but you can change its behavior by overriding the equals method in the classes you define. It is not possible to override == directly, as it is defined as

问题 I'm reading through Programming in Scala. It says: You can redefine the behavior of == for new types by overriding the equals method, which is always inherited from class Any . The inherited equals , which takes effect unless overridden, is object identity, as is the case in Java. So equals (and with it, == ) is by default the same as eq , but you can change its behavior by overriding the equals method in the classes you define. It is not possible to override == directly, as it is defined as

问题 In Java the following code can return false: IntegerPlus o1 = new IntegerPlus(1000); IntegerPlus o2 = o1; boolean b1 = o1 == o2; boolean b2 = o1.Equals (o2); Is this also a problem in C#? Or does C# perform the == in a way where it will always be true even if objects get moved? (I describe the Java issue in greater detail here.) 回答1: No. In C#/.Net hash code is not involved into default implementation of == , != or Equals . If object is reference type and it is moved by GC there is nothing

问题 Is there ever a situation where using equals(Boolean) and == would return different results when dealing with Boolean objects? Boolean.TRUE == myBoolean; Boolean.TRUE.equals(myBoolean); I'm not thinking about primitive types here, just Boolean objects. 回答1: How about: System.out.println(new Boolean(true) == new Boolean(true)); System.out.println(new Boolean(true) == Boolean.TRUE); (both print false, for the same reason as any other type of objects). 回答2: It would be dangerous to use ==

问题 Consider the following example: class Quirky { public static void main(String[] args) { int x = 1; int y = 3; System.out.println(x == (x = y)); // false x = 1; // reset System.out.println((x = y) == x); // true } } I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side ( x = y ) which, by the order implied by brackets, should be calculated first. Why does the first expression evaluate to

问题 This question already has answers here : Comparing Integer values in Java, strange behavior (3 answers) Closed 4 years ago . Integer comparison in Java is tricky, in that int and Integer behave differently. I get that part. But, as this example program shows, (Integer)400 (line #4) behaves differently than (Integer)5 (line #3) . Why is this?? import java.util.*; import java.lang.*; import java.io.*; class Ideone { public static void main (String[] args) throws java.lang.Exception { System.out