assignment-operator

问题 I thought the syntax: var a, b, c = {}; would mean that the three variables are separate, not references to the same {}. Is this because {} is an object and this is the standard behavior? So if I do: var a, b, c = 0; the three would indeed be separate and not references? Thanks, Wesley 回答1: They shouldn't be the same, no. Only c will be assigned the value. a and b would just be declared, but not initialized to anything (they'd be undefined). c would, as the only one of them, be initialized to

问题 This question already has answers here : In C, why can't I assign a string to a char array after it's declared? (6 answers) Closed 4 years ago . I tried doing this but got an error. Why can't I do this? int main() { char sweet[5]; sweet = "kova"; printf("My favorite sweet is %s\n", sweet); return 0; } 回答1: Nope, you cant. Simply because array names are non-modifiable l-value . Which cannot be used as left operand in any expression. Therefore, you cannot keep it on the left side of a =

问题 #include <stdio.h> int main() { char gradesList[5]; gradesList[2] = "X"; printf("%c", gradesList[2]); } When I try to run this code I get these errors: Incompatible pointer to integer conversion Assignment makes integer from pointer without a cast 回答1: You have to assign a char not a pointer to a string literal . Use ' instead of " gradesList[2] = 'X'; In C string literals are represented using double qoutes, i.e. " . And char are represented using single quotes, i.e. ' . Since you have

问题 This question already has answers here : Javascript object bracket notation ({ Navigation } =) on left side of assign (4 answers) Closed 2 years ago . How gonna assign in const { Types, Creators } in the below code I mean what Types gonna hold and what Creators gonna hold. const { Types, Creators } = createActions({ userRequest: ['username'], userSuccess: ['avatar'], userFailure: null }) var createActions = (function (config, options) { if (R.isNil(config)) { throw new Error('an object is

问题 I have a TemplateArray and a CharArray class. How do I make it so the templatearray's assignment operator only copies in from the chararray class when the templatearray is of the same type (I.E. char) or similar type (I.E. unsigned char) to chararray? TemplateArray and CharArray are functionally the same (except CharArray can handle NULL terminated strings). For example: template<typename TemplateItem> TemplateList & TemplateList<TemplateItem>::operator=(const CharArray &ItemCopy) { //How do

问题 Suppose I want to implement a class which is copyable, so I can implement the copy constructor and assignment operator. However, what is the correct implementation and handling of unique and shared pointer variables? See this contrived example which has both types of pointers: Header File #include <memory> using std::unique_ptr; using std::shared_ptr; class Copyable { private: unique_ptr<int> uniquePointer; shared_ptr<int> sharedPointer; public: Copyable(); Copyable(int value); Copyable(const

问题 Does the overloading of the assignment operator propagate to an initializer list? For example, suppose a class: class MyClass { private: std::string m_myString; //std::string overloads operator = public: MyClass(std::string myString); } And a constructor: MyClass::MyClass(std::string myString) : m_myString(myString) { } Will the initializer list work out the assignment operator overload on std::string ? And if not, is there a workaround? Particularly for GCC. 回答1: I think what you are missing

问题 I am using the ++: operator to get a collection of two collections, but the results I get using these two methods are inconsistent: scala> var r = Array(1, 2) r: Array[Int] = Array(1, 2) scala> r ++:= Array(3) scala> r res28: Array[Int] = Array(3, 1, 2) scala> Array(1, 2) ++: Array(3) res29: Array[Int] = Array(1, 2, 3) Why do the ++: and ++:= operators give different results? This kind of difference does not appear with the ++ operator. The version of Scala I am using is 2.11.8. 回答1: Since it

问题 For example: class Foo : public Bar { ~Foo() { // Do some complicated stuff } Foo &operator=(const Foo &rhs) { if (&rhs != this) { ~Foo(); // Is this safe? // Do more stuff } } } Are there any unexpected consequences of calling the destructor explicitly with regard to inheritance and other such things? Is there any reason to abstract out the destructor code into a void destruct() function and call that instead? 回答1: Calling the destructor is a bad idea in the simplest case, and a horrible one

问题 I have written a code, for dynamically allocating a name. I know I should take care of deep copy in such scenarios. What I have written is my own version of Copy Constructor,Copy Assignment Operator and destructor. Should I redefine any other implicit functions such as Move Assignment Operator . I am not clear with the concept of Move Assignment Operator or any other implicitly defined member functions (other than which I have already mentioned ). Can any one please add the code for this