assignment-operator

assignment operator String object

旧巷老猫 提交于 2019-12-02 08:04:16
I am new to JAVA programming. I have read it in my book String a="Hello"; String b="Hello"; System.out.println(a==b); This should return false as a & b refer to different instances of String objects. Bcoz the assignments operator compares the instances of objects but Still I am getting a true . I am using Eclipse IDE. Example in book goes as this: String s = "s"; String sToo = "s"; System.out.println(a == b); System.out.println(s == sToo); That bit of code prints “false” for s == sToo. That's because s and sToo are references to different instances of the String object. So, even though they

How can I initialize the default value of a CArray<CClass*> function parameter with an empty CArray?

我是研究僧i 提交于 2019-12-02 07:36:13
I know I could do this better with std::vector , but the application I am messing with, has already a bunch of CArray parameters on a lot of related functions ... and I will not change them all at the moment! I simply want to define an empty CArray<CClass*> — array of pointers to CClass , so the problem can not be on the CClass constructor — as the default value of a function parameter. Approach 1 If I try to do it with assignment operator and default constructor : void Function(CArray<CClass*> parameter = CArray<CClass*>()); I get the error: 1>C:\Program Files (x86)\Microsoft Visual Studio 10

Relationship between assignment operator and user-defined constructor

北城以北 提交于 2019-12-02 06:05:00
问题 #include <iostream> class A{ }; class B: public A{ public: B(A&& inA){ std::cout<<"constructor"<<std::endl; } }; int main(){ B whatever{A{}}; whatever=A{}; } This outputs constructor constructor at least with C++14 standard and GCC. How is it defined that assignment operator can result in call to constructor instead of operator= ? Is there a name for this property of assignment operator? 回答1: Since you meet all the conditions for generating a move-assignment operator. The move-assignment

Double-assignment of an object property results in undefined property [duplicate]

狂风中的少年 提交于 2019-12-02 04:51:42
This question already has an answer here: Assignment associativity [duplicate] 1 answer Can anyone tell how the output became undefined ? var foo = {n: 2}; foo.x = foo = {n: 2}; console.log(foo.x); // undefined foo.x = foo = {n:2}; The foo.x refers to the property x to the object refered to by foo . However, foo = {n:2} assigns a completely new object to foo . x is indeed assigned to an object, but that object is immediately replaced by another object. The object with the x property isn’t referenced by anything anymore. You can read that line as foo.x = (foo = {n:2}); Graphical explanation var

Assigning *&array to a pointer

試著忘記壹切 提交于 2019-12-02 04:23:52
问题 The following excerpt is from Harbinson, Steele C: A Reference Manual (5th Edition). According to the book the two assignments to p are equivalent. 7.5.6 Address Operator int a[10], *p; p = a; p = *&a; Yet, according to the C faq Question 6.12 a is of type pointer to int whereas &a is of type pointer to array of int . So we should get a type error in the second assignment p = *&a because we are trying to assign an array of int to a pointer. Why is the assignment p = *&a correct? 回答1: Quoting

Assignment of a Singular Iterator

柔情痞子 提交于 2019-12-02 03:37:40
A "Singular Iterator" is defined as an: iterators that are not associated with any sequence. A null pointer, as well as a default-constructed pointer (holding an indeterminate value) is singular My question 1 would be: Is a default constructed iterator considered a "Singular Iterator"? Secondly, I have been told here that: Results of most expressions are undefined for singular values; the only exceptions are destroying an iterator that holds a singular value, the assignment of a non-singular value to an iterator that holds a singular value, and, for iterators that satisfy the

behaviour of the implicit copy constructor / assignment operator

蓝咒 提交于 2019-12-02 01:41:19
问题 I have a question regarding the C++ Standard. Suppose you have a base class with user defined copy constructor and assignment operator. The derived class uses the implicit one generated by the compiler. Does copying / assignment of the derived class call the user defined copy constructor / assignment operator? Or do you need to implement user defined versions that call the base class? Thank you for your help. 回答1: If a derived class does not declare a copy constructor, and implicit one will

behaviour of the implicit copy constructor / assignment operator

余生颓废 提交于 2019-12-01 23:58:18
I have a question regarding the C++ Standard. Suppose you have a base class with user defined copy constructor and assignment operator. The derived class uses the implicit one generated by the compiler. Does copying / assignment of the derived class call the user defined copy constructor / assignment operator? Or do you need to implement user defined versions that call the base class? Thank you for your help. If a derived class does not declare a copy constructor, and implicit one will be declared (12.8/4 "Copying class objects") - even if the base class has a user-delcared and defined copy

derived class's virtual assignment operator not being called

◇◆丶佛笑我妖孽 提交于 2019-12-01 22:19:24
I'm pretty new to C++, and am trying to come to grips with virtual assignment. The program below consists of an abstract base class with two data members, and a derived class with one. When I set an abstract pointer to a derived object, the program uses the abstract version of operator= rather than the derived version, even though they're both declared "virtual." What am I doing wrong here? Thanks in advance, Jay #include <iostream> #include <cstring> class Abstract { protected: char * label; int rating; public: Abstract(const char * l = "null", int r = 0); virtual Abstract & operator=(const

“error: lvalue required as left operand of assignment” in conditional operator

无人久伴 提交于 2019-12-01 19:54:46
问题 I'm new to C and today I learnt "?" operator which is the short type of if-else statement. However, when I execute this code: int b; int x; b=3<2?x=12:x=34; I get an error "error: lvalue required as left operand of assignment". I don't understand why it happens. Process in my mind is that the program first assigns 34 to x, then it assigns value of x,which is 34, to b. On the other hand, I can use the statement as int b; int x; b=3<2?x=12:(x=34); without any errors. I looked to my book but