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What is the valid class declaration header for the derived class d with base classes b1 and b2?A. class d : public b1, public b2 {/*...*/};B. class d : class b1, class b2 {/*...*/};C. class d : public b1, b2 {/*...*/};D. class d : b1, b2 {/*...*/};The answer is A, C and D.B is not valid because "class" is not a valid access specifier.All the others are valid because private access is the default when the access specifier is omitted. Note that if class D were declared using the struct prefix, inheritance would default to public access rather than private.
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Multiple inheritance occurs when a class is derived directly from two or more base classes. class b1 {}; class b2 {}; class d: public b1, public b2 {}; // multiple inheritance class
Whenever a derived class requires direct inheritance from a base class, even if it inherits that base class indirectly. That is, if V is a base class from which B is derived, and D is derived from B, then D inherits from V indirectly (through B). But if B is virtually derived from V, then D will inherit directly from V. This feature is commonly used in conjunction with multiple inheritance. Examine the following declarations: class V{}; class B1: public V{}; class B2: public V{}; class M: public B1, public B2{}; Now suppose you have the following code: M m; // Declare an instance of M. V& v = m; // Ambiguous... The problem with this is that M inherits V from both B1 and B2, and therefore inherits two separate instances of V. The compiler is unable to determine which instance of V you want to refer to. One solution to this would be to use static casts to indirectly refer to an explicit instance of V: V& v = static_cast<B1&>(m); or V& v = static_cast<B2&>(m); While this is certainly workable, it is an ugly approach that places far too much responsibility upon the programmer to ensure the correct instance of V is being referred to. However, unless there is a specific need to have two instances of V within M, the problem can be resolved with virtual inheritance. By virtually deriving both B1 and B2 from V, M will directly inherit just one instance of V, which is then shared, virtually, between B1 and B2: class V{}; class B1: public virtual V{}; class B2: public virtual V{}; class M: public B1, public B2{}; M m; V& v = m; // No ambiguity. Now M can access all the members of V directly, as can B1 and B2, because they now share the same instance of V. Note that it doesn't matter whether the virtual keyword is placed before or after the access specifier (which is public in this case). "virtual public" and "public virtual" have the same meaning.
Object slicing typically occurs when operating upon a base class reference, even when the reference is actually an instance of a derived class. You might expect polymorphic behaviour but it doesn't happen, due to object slicing. Most of the time this is only to be expected (such as when passing a derived object to a base class copy constructor or assignment operator), but when you actually expect polymorphic behaviour it can catch you out. The following example of partial assignment demonstrates this:#includestruct A {int a_var;A(int a): a_var(a) {}void debug() { std::cout
There are an infinite number of solutions. Blank1 = 1 and blank2 = 15 B1 = -1 and B2 = -15 B1 = 2 and B2 = 7.5 B1 = 3 and B2 = 5 B1 = 1.1 and B2 = 15/1.1 etc, etc.
You could use either of the following, by putting the formulas in any cells except A1 and B1: =A1+B1 =SUM(A1:B1)
What is the valid class declaration header for the derived class d with base classes b1 and b2?A. class d : public b1, public b2 {/*...*/};B. class d : class b1, class b2 {/*...*/};C. class d : public b1, b2 {/*...*/};D. class d : b1, b2 {/*...*/};The answer is A, C and D.B is not valid because "class" is not a valid access specifier.All the others are valid because private access is the default when the access specifier is omitted. Note that if class D were declared using the struct prefix, inheritance would default to public access rather than private.
t1 s1 b1 t1 s1 b2 t1 s2 b1 t1 s2 b2 t2 s1 b1 t2 s1 b2 t2 s2 b1 t2 s2 b2 t3 s1 b1 t3 s1 b2 t3 s2 b1 t3 s2 b2 TOTAL 12 combinations OR 2(3 x 2) = 12
Perimeter = area + b1 + b2 + c P = a + b1 + b2 + c
The equation A H b1 plus b2 times 2 equals 8. This is a math problem.
b1 and b2
The b-2 has about twice the capacity of a b-1.
AREA = A = 1/2 (b1 + b2)(h) h= height = 2(A) / (b1 + b2)
start at Ic'= Ic1+Ic2 Ib'=Ib1 Ic1=B1*Ib' Ie1=Ib2=Ib'(B1+1)=Ib'B1 + Ib' Ic' = Ib'*B1 + Ic2 Ic2= B2(Ib'B1+Ib') Ic' = Ib'B1 + Ib'B1B2 + Ib'B2 divide by Ib' Ic'/Ib'= current gain =B'=B1+B2+B1B2 note: calculations without resistance
You have 4 6V batteries. Let's name them B1, B2, B3, B4. Now, let's name the terminals on them, so B1 has B1- and B1+ Now let's name your final output of 24V O- and O+ Connect it like this: B1- to O- B1+ to B2- B2+ to B3- B3+ to B4- B4+ to O+
CLass 1 Class 2 class 3 16-26ft 26 ft-40 ft 40ft-65ft 1 B1 2 b1 or 1 b2 3 b1 or 1 b1 and 1 B2 Im taking a boating class currently