Since Test is within namespace B the compile sees the operator in that namespace and notes that it doesn’t have a matching signature. It also attempts to find the operator in namespace A which contains the class but can’t find it there either. Because there’s already such an operator (with the wrong signature) in namespace … Read more
From the Spec: 6.12.3 InfixOperations An infix operator can be an arbitrary identifier. Infix operators have precedence and associativity defined as follows. … The associativity of an operator is determined by the operator’s last character. Operators ending in a colon ‘:’ are right-associative. All other operators are left- associative. You can always see how these … Read more
The key to the answer is called object interning. Java interns small numbers (less than 128), so all instances of Integer(n) with n in the interned range are the same. Numbers greater than or equal to 128 are not interned, hence Integer(1000) objects are not equal to each other.
Prior to C++23, you could not overload operator[] to accept multiple arguments. As a workaround, you instead can overload operator(). (See How do I create a subscript operator for a Matrix class? from the C++ FAQ.) From C++23, as mentioned in a (deleted) answer by cigien, multiple subscript arguments can be passed to operator[] directly. … Read more
Your code, as written, works. You’re probably trying to achieve something unrelated, but similar: std::string c = “hello” + “world”; This doesn’t work because for C++ this seems like you’re trying to add two char pointers. Instead, you need to convert at least one of the char* literals to a std::string. Either you can do … 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
Not out of the box. However, it’s easy to build by hand in many languages including JS. var operators = { ‘+’: function(a, b) { return a + b }, ‘<‘: function(a, b) { return a < b }, // … }; var op = ‘+’; alert(operators[op](10, 20)); You can use ascii-based names like plus, … Read more
Looks like you’re a little bit confused about all that stuff. operator is a built-in module providing a set of convenient operators. In two words operator.itemgetter(n) constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it. So, you can’t use key=a[x][1] there, because … Read more
The >> operator in your example is used for two different purposes. In C++ terms, this operator is overloaded. In the first example, it is used as a bitwise operator (right shift), 2 << 5 # shift left by 5 bits # 0b10 -> 0b1000000 1000 >> 2 # shift right by 2 bits # … Read more
Use object.ReferenceEquals(person1, null) or the new is operator instead of the == operator: public static bool operator ==(Person person1, Person person2) { if (person1 is null) { return person2 is null; } return person1.Equals(person2); }