Are variadic constructors supposed to hide the implicitly generated ones, i.e. the default constructor and the copy constructor?
Suppose I have a class that requires copy constructor to be called to make a correct copy of: struct CWeird
I know that Qobjects are supposed to be identities not values eg you cannot copy them and by default the copy constructor and assignment are disabled as explained in qt documentation. But is it possib
I have this code #include <iostream> using namespace std; class Test{ public: int a; Test(int i=0):a(i){}
I was as开发者_开发百科ked this question in one of my interviews, but I wasn\'t able to find out exactly why this concept is not there.
[a follow up to this question] class A { public: A(){cout<<\"A Construction\"<<endl;} A(A const& a){cout<<\"A Copy Construction\"<<endl;}
Stepping through my program in gdb, line 108 returns right back to the calling function, and doesn\'t call the copy constructor in class A, like (I thought) it should:
I have two copy constructors Foo(Foo &o开发者_开发百科bj){ } Foo(Foo *obj){ } When will the second copy constructor will get called?No you don\'t - you have one copy constructor
Here is an extract from item 56 of the book \"C++ Gotchas\": It\'s not uncommon to see a simple initialization of a Y object written
What is \"minimal framework\" (necessary methods) of complex object (with explicitly malloced internal data), which I want to store in STL container, e.g. <vector>?
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