Let’s say MVP didn’t exist. How would you declare a function? A foo(); would be a variable definition, not a method declaration. Would you introduce a new keyword? Would you have a more awkward syntax for a function declaration? Or would you rather have A foo; define a variable and A foo(); declare a function? … Read more
Clang gives this warning message: <source>:12:16: warning: parentheses were disambiguated as redundant parentheses around declaration of variable named ‘num’ [-Wvexing-parse] Boo(num); // No default constructor ^~~~~ This is a most-vexing parse issue. Because Boo is the name of a class type and num is not a type name, Boo(num); could be either the construction of … Read more
nothing at all, because X(answer); could be interpreted as the declaration of a variable. Your answer is hidden in here. If you declare a variable, you invoke its default ctor (if non-POD and all that stuff). On your edit: To get a temporary, you have a few options: (X(answer)); (X)answer; static_cast<X>(answer) X{answer}; (C++11) []{ return … Read more
It’s interpreted as the declaration of a function named a, which takes one argument of type B and returns A.
Grouping. As a particular example, consider that you can declare a variable of function type such as int f(int); Now, how would you declare a pointer to such a thing? int *f(int); Nope, doesn’t work! This is interpreted as a function returning int*. You need to add in the parentheses to make it parse the … Read more
Most vexing parse This is related to what is known as “C++’s most vexing parse”. Basically, anything that can be interpreted by the compiler as a function declaration will be interpreted as a function declaration. Another instance of the same problem: std::ifstream ifs(“file.txt”); std::vector<T> v(std::istream_iterator<T>(ifs), std::istream_iterator<T>()); v is interpreted as a declaration of function with … Read more