c Answer to the exercise

c is defined outside of class, so it is a global variable. A is a private variable defined in class a.

Here c is a global variable, b is a static variable (static), and a is a normal variable. b is initialized to 0

These set functions are to add 1 to the corresponding variable

A, B, and C in A1 are 1 0 0, respectively

a1 1 1 0

a1 1 1 1

A, B, and C in A2 are 1 1 1 respectively (B is static and remains 1).

a2 1 2 1

a2 1 2 2

So; Output 1 2 2

while(at is a variable that temporarily saves values t=a; a=b;b=t;It's a and B swapped.

c -- is c minus 1

The program runs the first while judgment a and then a, b, and c become a=2, b=1, c=1

The second while judgment a then a, b, c becomes a=1, b=2 , c=0

The third while judgment a outputs the values of a,b,c 1,2,0

str is the name of the array, which is a pointer to the head of the array.

str+2 is the position of the last two characters at the beginning of the array str, so the z is covered.

The end result str is xyabcabc

Alas......These questions are the most annoying for the examiners.

1.A is defined inside the class, C is defined outside the class!! A is the class name, A1, A2 objects, is the call of the function!!

2.If while(a is the third character in str!)

I'm also a beginner, so maybe there's something wrong with what I've said, for reference!

1.The variables in the class will be added with a this pointer, and a will become this ->a, and c is not in the class a, so this pointer is not added (this pointer will be automatically passed in when called, without manual intervention).

Then, b is a static property, which does not belong to a particular object, and belongs to the class a, int a::b=0; This sentence opened up memory space for him and initialized it, and then the two setb operations were the same B.

As for those formulas, they call the methods in the A1 object and the A2 object.

a::b = 1c = 1

a::b = 2

c = 22.The problem lies in Aa = 2 b = 1 c = 1 a < b the result is false i.e. 0, 0 < c the result is true, continue.

a = 1 b = 2 c = 0 a < b the result is true i.e. 1, 1 < c the result is false, end.

The pointer is useful because you can add or subtract it directly, and +1 will move to the next one (regardless of the type), such as int a[10], a points to a[0], a + 1 points to a[1], a + 2 points to a[2], and the same goes for char arrays, str points to str[0], str + 2 points to str[2].

Attention, str + 2 is an address!

Q1: First of all, it is clear that a is defined in class a, which is an ordinary variable. c is defined outside of the function so it is a global variable, it does not belong to any function, it belongs to a source program file. Its scope is the entire source program.

Secondly, the main function first defines the two objects A1 and B1 in class A, and then accesses the member functions in the class objects in turn. Get a=1;

Get b=1;c=1;

Get a=1;(Because there is int a::b=0; b is a local variable acting on class a, so the initial b = 1 in a2 and the initialization of a in a2 is 0; c is a global variable, c=1).

Get b=2;c=2;

Hence the output 1 2 22 q.