No, you can’t use explicit, but you can use a templated function to catch the incorrect parameter types. With C++11, you can declare the templated function as deleted. Here is a simple example: #include <iostream> struct Thing { void Foo(int value) { std::cout << “Foo: value” << std::endl; } template <typename T> void Foo(T value) … Read more
The implicit and explicit keywords in C# are used when declaring conversion operators. Let’s say that you have the following class: public class Role { public string Name { get; set; } } If you want to create a new Role and assign a Name to it, you will typically do it like this: Role … Read more
Yes and No. It depends on which version of C++, you’re using. C++98 and C++03 do not support explicit type conversion operators But C++11 does. Example, struct A { //implicit conversion to int operator int() { return 100; } //explicit conversion to std::string explicit operator std::string() { return “explicit”; } }; int main() { A … Read more
The rows you get back from iterrows are copies that are no longer connected to the original data frame, so edits don’t change your dataframe. Thankfully, because each item you get back from iterrows contains the current index, you can use that to access and edit the relevant row of the dataframe: for index, row … Read more
Indeed, very interesting behavior. Any idea why I get I runtime error when commenting out #include <string> With MS VC++ compiler the error happens because if you do not #include <string> you won’t have operator<< defined for std::string. When the compiler tries to compile ausgabe << f.getName(); it looks for an operator<< defined for std::string. … Read more
The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter. Here’s an example class with a constructor … Read more