Some time ago I was told, that the usual pattern to implement two-ary operators needs a final move in the return.
I have a doubt about how C++ casts types when it has to do math. The code below as it is ( i.e. with only the cast to int without casting to double ) works and builds without problem. If I define EN
Consider the following constructors for T: struct T { T(const bool) { std::cout << \"T(const bool)\" << endl; }
Code: class que { public: que operator++(int) {}// 1 que &operator++() {} que &operator+=(int n) {
This question already has answers here: Closed 11 years ago. Possible Duplicate: Why must we define both == and != in C#?
I want to overload (hijack?) ostream and basic_ostream<unsigned char> so that it stops attempting to display an octet (unsigned char) as a printable character.
I need an in implicit conversion from A* to C*; i cannot change A\'s definition or implementation. class A
i wrote a class called integer that can handle integers of arbitrary bit sizes, and i seem to have completed everything except one thing, which despite everything that i do to try to fix it, refuses t
Can a working factorial function/operator be defined with a syntax like in mathematics?i.e. using the ! symbol.
I have an object that has a mathematical function behind it.It seems like a perfect candidate for operator().
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