operator-keyword

问题 could somebody explain the difference between these two codes? bad_command 2>& >> file.out and bad_command >> file.out 2>& The manual said that these two codes are different,and first command will output nothing to file.out. So , here are my questions. 1/ what is the reason for that? 2/ Is there is a document which describes how operator precedence works in shell? how shell parses and made it's syntax tree. 3/ What is the correct syntax and order of it? --Thanks in advance-- 回答1: Both are

问题 I'm just starting out in AP Comp sci in high school and I stumbled across a question regarding the + operator in strings Why does System.out.println ("number" + 6 + 4 * 5) result in number620 whereas String s = "crunch"; int a = 3, b = 1; System.out.print(s + a + b); System.out.print(b + a + s); result in crunch314crunch ? Thanks 回答1: Depends on It's precedence order When two operators share an operand the operator with the higher precedence goes first. For example, 1 + 2 * 3 is treated as 1

问题 For a programme that I must write for an assignment. I've been told to use either a Smart Match Operator or Hashes. My problem is, is that I just don't understand how they work or what they do no matter what I read in textbooks or on the internet. I don't suppose somebody could explain their function to me really simply? The final part of the assignment is to deduce the number of unique species in an area - I've calculated which species are the in the area but I don't know how to calculate

问题 Consider: operator.add(a, b) I'm having trouble understanding what this does. An operator is something like +-*/ , so what does operator.add(a, b) do and how would you use it in a program? 回答1: Operator functions let you pick operations dynamically . They do the same thing as the operator, so operator.add(a, b) does the exact same thing as a + b , but you can now use these operators in abstract. Take for example: import operator, random ops = [operator.add, operator.sub] print(random.choice

问题 I was previously under the (naive) assumption that the modulus operator returned the remainder of division. I was apparently wrong, as -2 % 5 returns 3. I would have thought that 5 divides -2 zero times with -2 as the remainder. Now I understand the mechanics of how this operation is performed, but my question is why? Could someone give me a link to something that explains why modulus and remainder are not synonymous, or an example of a situation where it would be useful? 回答1: The result is

问题 I have never done any extensive work with overloading operators, especially the implicit and explicit conversions. However, I have several numeric parameters that are used frequently, so I am creating a struct as a wrapper around a numeric type to strongly type these parameters. Here's an example implementation: public struct Parameter { private Byte _value; public Byte Value { get { return _value; } } public Parameter(Byte value) { _value = value; } // other methods (GetHashCode, Equals,

问题 This question already has answers here : How do I override the bool operator in a C++ class? (4 answers) Closed 6 years ago . For example, I have defined a class class MyClass { .... }; which operator do I have to overload in order to do the if comparison with a MyClass object? eg: MyClass cc; if ( cc ) // compile error { //do some stuff } I tried with bool operator == ( int value ) ; //guess the if () will call this or bool operator != ( int value ) ; but both give me a compile error! 回答1:

问题 I'm using GNAT GPS studio IDE in order to train a bit in Ada. I'm having an issue with package visibility. First I specify a package in a file called "DScale.ads" containing a type: package DScale is type DMajor is (D, E, F_Sharp, G, A, B, C_Sharp); end DScale; Then I specify in a different file ("Noteworthy.ads") a package that defines a procedure that will use the DMajor type of the DScale package: with Ada.Text_IO; with DScale; package NoteWorthy is procedure Note; end NoteWorthy; Finally

问题 Is there a meaning to declare friendship in the protected section, rather than in public? For example in this code: class Shape { //... protected: friend ostream& operator<<(ostream& os, const Shape& s); virtual void print(ostream& os) const = 0; }; [Note that Shape is abstract] Could I have just put the friend and the function declaration in public? Thanks! 回答1: Is there a meaning to declare friendship in the protected section, rather than in public? No. The friend class has the same level

问题 Here is an example from website: http://www.cplusplus.com/doc/tutorial/classes2/ I know it is a working example. However, I don't understand why object temp can be returned from the operator+ overloading function. I have made some comments besides the codes. // vectors: overloading operators example #include <iostream> using namespace std; class CVector { public: int x,y; CVector () {}; CVector (int,int); CVector operator + (CVector); }; CVector::CVector (int a, int b) { x = a; y = b; }