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This appears to be undefined behavior
union A { int const x; float y; }; A a = { 0 }; a.y = 1; The spec says
Creating a new object at the storage location that a const object with static, thread, or automatic storage duration occupies or, at the storage location that such a const object used to occupy before its lifetime ended results in undefined behavior.
But no compiler warns me while it's an easy to diagnose mistake. Am I misinterpreting the wording?
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A union can have member functions (including constructors and destructors), but not virtual functions. A union cannot have base classes and cannot be used as a base class. A union cannot have non-static data members of reference types.
In C++17 and later, the std::variant class is a type-safe alternative for a union. A union is a user-defined type in which all members share the same memory location. This definition means that at any given time, a union can contain no more than one object from its list of members.
union { int no; char name[20], float price; }; In this, the size of the member, name is more than the size of other members (sr, price) so the size of a union in memory (char × 20 = 20 byte) will be 20 bytes.
The latest C++0x draft standard is explicit about this:
In a union, at most one of the non-static data members can be active at any time, that is, the value of at most one of the non-static data members can be stored in a union at any time.
So your statement
a.y = 1; is fine, because it changes the active member from x to y. If you subsequently referenced a.x as an rvalue, the behaviour would be undefined:
cout << a.x << endl ; // Undefined! Your quote from the spec is not relevant here, because you are not creating any new object.
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