operator-keyword

问题 This leads to the situation like: -1 ** 0.5 #=> -1 Only parenthesis remedies it: (-1) ** 0.5 #=> 6.123031769111886e-17+1.0i which is less favorable then expected 1.i , but basically acceptable. Before I go to Ruby bugs to complain, I would like to know whether there is perhaps some reason for this to be so? 回答1: Many languages define their operator precedence tables by modeling after mathematics' order of operations. In math, exponentiation does have higher precedence than multiplication, and

问题 I'm learning about ternary operators now. I got the basics down, but then I saw this snippet and it doesn't make any sense to me. Can anyone please explain how is it put together?! b.m & 4 || (c |= 2, 63 <= a && 77 >= a ? a = 65 : 48 <= a && 57 >= a ? a = 48 : c & 1 ? 97 <= a && 122 >= a ? a = 65 : 197 == a || 229 == a ? c &= 5 : 192 <= a && 687 >= a ? a = 192 : 1536 <= a ? a = 1536 : 912 <= a ? a = 912 : 160 <= a ? a = 160 : 127 <= a ? c &= 5 : 33 <= a ? a = 59 : c &= 5 : 48 > a ? c &= 5 :

问题 This is the smallest contained example I can think of. First the header of the class. This class should simply print the one double it contains whenever the << operator is used. #pragma once #ifndef EURO_H #define EURO_H #include <ostream> namespace EU { class Euro final { public: explicit Euro(double value); virtual ~Euro() = default; double getValue() const; friend std::ostream& operator<<(std::ostream &os, const Euro &euro); private: double m_value; }; } #endif // EURO_H Now the .cpp

问题 I got a Security Class that has an array of Predictions - Prediction is a Class, which holds only a double. I want to allow changing the value of the double, but allow only positive values, and when trying to read the double, if the value is uninitialized (equals -1 in my code) throw exception. I also have double operator in Something like that: class Prediction{ double value; public: ..... Prediction::operator double() const { return this->prediction; } Prediction::operator=(const double

问题 C++ FAQ is defining a template container Matrix to avoid tricky new delete code. Tutorial says that subscript operators often come in pairs ? Why is it so ? T& operator() (unsigned i, unsigned j); T const& operator() (unsigned i, unsigned j) const; Why is it so ? This is also called : const-overloading. FAQ gives clues. Do you have additional comments ? In particular, should mutate() observe certain rules to be used safely on const objects only ? 回答1: To put it simply, because there are two

问题 What is the fixity of backtick operators? For instance in this code from Real World Haskell: ghci> (1+) `fmap` [1,2,3] ++ [4,5,6] [2,3,4,4,5,6] It's evident the backtick operator `fmap` has a higher fixity than ++ , but none is given by GHCi. 回答1: §4.4.2 of the Haskell Report states that Any operator lacking a fixity declaration is assumed to be infixl 9 "Any operator" includes normal function names in backticks. Your example shows that `fmap` does have higher fixity than ++ , because ++ acts

问题 i have code like this: var db: name = dbFunction(true); dbFunction returning Object. I have question, what doing this colon operator in variable name? 回答1: It's a high tech operator that guarantees a syntax error when used like that. In it's normal use, you might see it used in object literal syntax to denote key:value pairs; var object = { "name": "value", "name2": "value2" } It can also be used to define a label (less common). loop1: for (var i=0;i<10; i++) { for (var j=0;j<10;j++) { break

问题 While porting Windows code to Linux, I encountered the following error message with GCC 4.2.3. (Yes, I'm aware that it's a slight old version, but I can't easily upgrade.) main.cpp:16: error: call of overloaded ‘list(MyClass&)’ is ambiguous /usr/include/c++/4.2/bits/stl_list.h:495: note: candidates are: std::list<_Tp, _Alloc>::list(const std::list<_Tp, _Alloc>&) [with _Tp = unsigned char, _Alloc = std::allocator<unsigned char>] /usr/include/c++/4.2/bits/stl_list.h:484: note: std::list<_Tp,

问题 I'm writing a template matrix class, and I am a bit confused as to how overloading the * operator would work. I would want to overload something like this (omitting irrelevant code): template<typename T>class Matrix4t { friend vector3 operator*(const vector3 &inputV3, const matrix4t &inputM4t); template <typename scalarT> friend scalarT operator*(const scalarT &inputSclT, const matrix4t &inputM4t); public: const matrix4t operator*(const matrix4t) vector3 operator*(const vector3 &inputV3);

问题 What is the difference between #= and =:= in SWI prolog. I have found the definition from SWI prolog, but still confused about it. http://www.swi-prolog.org/pldoc/man?section=arithpreds http://www.swi-prolog.org/pldoc/man?section=clpfd-arith-constraints ?- 3=:=3. true. ?- (3-2) =:= (9-8). true. ?- 3 #= 3. true. ?- (3-2) #= (9-8). true. 回答1: What's the difference between #= and =:= in SWI prolog ? The difference is that #=/2 is a CLPFD library operator (you need to execute: use_module(library