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The following code compiles fine with both gcc 7.2.0 and clang 6.0.0.
#include <iostream>
struct stru;
void func(stru& s) {
std::cout << &s << std::endl;
}
int main() {
}
I'm wondering how this is OK. What if stru has overloaded operator&()? The compiler should not be able to tell with simply a forward declaration like struct stru. In my opinion, only std::addressof(s) is OK with an incomplete type.
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The unary operators require only one operand; they perform various operations such as incrementing/decrementing a value by one, negating an expression, or inverting the value of a boolean. The increment/decrement operators can be applied before (prefix) or after (postfix) the operand.
Unary Increment (++) In this type, the Unary Operator is denoted by the '++' sign. It increases the value by 1 to the operand and then stores the value to the variable. It works for both Prefix and Postfix positions.
In mathematics, an unary operation is an operation with only one operand, i.e. a single input. This is in contrast to binary operations, which use two operands. An example is any function f : A → A, where A is a set. The function f is a unary operation on A.
What Does Unary Operator Mean? A unary operator, in C#, is an operator that takes a single operand in an expression or a statement.
What if
struhas overloadedoperator&()?
Then it is unspecified whether the overload will be called (See Oliv's comment for standard quote).
How could unary operator & does not require a complete type?
That's how the standard has defined the language. The built-in address-of operator doesn't need to know the definition of the type, since that has no effect on where to get the address of the object.
One consideration for why it is a good thing: Compatibility with C.
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