operator-keyword | 易学教程

How != and == operators work on Integers in Java? [duplicate]

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This question already has an answer here: Why is 128==128 false but 127==127 is true when comparing Integer wrappers in Java? 6 answers The following code seemed really confusing to me since it provided two different outputs.The code was tested on jdk 1.7. public class NotEq { public static void main(String[] args) { ver1(); System.out.println(); ver2(); } public static void ver1() { Integer a = 128; Integer b = 128; if (a == b) { System.out.println("Equal Object"); } if (a != b) { System.out.println("Different objects"); } if (a.equals(b)) { System.out.println("Meaningfully equal."); } }

Concatenating strings doesn't work as expected [closed]

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问题 I know it is a common issue, but looking for references and other material I don't find a clear answer to this question. Consider the following code: #include <string> // ... // in a method std::string a = "Hello "; std::string b = "World"; std::string c = a + b; The compiler tells me it cannot find an overloaded operator for char[dim] . Does it mean that in the string there is not a + operator? But in several examples there is a situation like this one. If this is not the correct way to

Haskell operator vs function precedence

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问题 I am trying to verify something for myself about operator and function precedence in Haskell. For instance, the following code list = map foo $ xs can be rewritten as list = (map foo) $ (xs) and will eventually be list = map foo xs My question used to be, why the first formulation would not be rewritten as list = (map foo $) xs since function precedence is always higher than operator precedence, but I think that I have found the answer: operators are simply not allowed to be arguments of

Namespaces and operator resolution

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问题 I am using a library that defines output stream operators (operator<<) in the global namespace. In my own namespace, I was always declaring such operators in the global namespace, and never had problems with them. But now for various reasons I need to declare those operators in my own namespace, and suddenly, the compiler can\'t seem to find the operators declared in the library anymore. Here\'s a simple example that illustrates my problem: #include <iostream> namespace A { struct MyClass {};

Chaining “is” operators

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问题 Does python support chaining is operators, such as the following? a = None b = None a is b is None This outputs True , some doc references would be nice. 回答1: Yes. Any operators classified as comparisons can be chained. From the language reference: Formally, if a , b , c , ..., y , z are expressions and op1 , op2 , ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z , except that each expression is evaluated at most once. The

Python's Logical Operator AND

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I'm a little confused with the results I'm getting with the logical operators in Python. I'm a beginner and studying with the use of a few books, but they don't explain in as much detail as I'd like. here is my own code: five = 5 two = 2 print five and two >> 2 It seems to be just outputting the two variable. five = 5 two = 2 zero = 0 print five and two and zero So, I added another variable integer. Then I printed and got the following output: >> 0 What is going on with Python in the background? Why isn't the output something like 7 or 5, 2. tdelaney Python Boolean operators return the last

Using == operator in Java to compare wrapper objects

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I'm reading SCJP Java 6 by Kathy Sierra and Bert Bates and this book is confusing me so much. On page 245 they state that the following code below. Integer i1 = 1000; Integer i2 = 1000; if(i1 != i2) System.out.println("different objects"); //Prints output different objects Then on the very next page they have the following code Integer i3 = 10; Integer i4 = 10; if(i3 == i4) System.out.println("same objects"); //Prints output same objects I'm so confused! When I try this out on my own it seems that you cannot use the == to compare the same way you would use equals() method. Using the == always

Java autoboxing and ternary operator madness

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问题 Just spent a frustrating couple of hours debugging this code: LinkedHashMap<String, Integer> rsrqs = new LinkedHashMap<String, Integer>(); Integer boxedPci = 52; Integer boxedRsrq = boxedPci != null ? rsrqs.get(boxedPci.toString()) : -1; The above produces a NullPointerException. The below code doesn\'t: LinkedHashMap<String, Integer> rsrqs = new LinkedHashMap<String, Integer>(); Integer boxedPci = 52; Integer boxedRsrq = boxedPci != null ? rsrqs.get(boxedPci.toString()) : Integer.valueOf(-1)

Can a cast operator be explicit?

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问题 When it comes to constructors, adding the keyword explicit prevents an enthusiastic compiler from creating an object when it was not the programmer’s first intention. Is such mechanism available for casting operators too? struct Foo { operator std::string() const; }; Here, for instance, I would like to be able to cast Foo into a std::string , but I don’t want such cast to happen implicitly. 回答1: Yes and No. It depends on which version of C++, you're using. C++98 and C++03 do not support

Scala's '::' operator, how does it work?

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问题 In Scala, I can make a caseclass, case class Foo(x:Int) , and then put it in a list like so: List(Foo(42)) Now, nothing strange here. The following is strange to me. The operator :: is a function on a list, right? With any function with one argument in Scala, I can call it with infix notation. An example is 1 + 2 is a function (+) on the object Int . The class Foo I just defined does not have the :: operator, so how is the following possible? Foo(40) :: List(Foo(2)) In Scala 2.8 RC1, I get