2020年电大面向对象程序设计期末复习指导参考必考重点【精编版】.

1、电大面向对象程序设计期末复习指导小抄一、单顼选题1.字符串a+b=12n”的长度为(A. 6B. 7)C.8D.92.x0| y=5的相反表达式为( A. x0 | y!=53.4.B. x0 & y=5循环体有可能一次都不执行的语句为(A.for 循环B. switch 循环C. do循环D.任一种循环函数调用func(exp1,exp2,exp3),exp4,exp5)中所含实参的个数为(A. 1B.2C. 3D.5)个。5.假定p是具有int，类型的指针变量，则给p赋值的正确语句为( A. p=new int;C. p=new int*;)。6.a34=(3,4,(2,8,6;A. 2B

2、.4B. p=new int10);D. p=new int*;假定一个二维数组的定义语句为 ，则元素a的值为()。C.6D.0int假定一个类的构造函数为 A(int aa=1, int bb=O) (a=aa; b=bb;,则执行A7.x(0);”语句后,x.a和x.b的值分别为()。A. 0 和 08.假定AA为一个类，inta()为该类的一个成员函数.若该成员函数在类定义体外定义，则函数头为(A . int AA:a()B . int AA:a()C . AA:a()D . AA:int a()9 .关于插入运算符 的重载.下列说法不正确的是(A. 运算符函数的返回值类型是ostrea

3、m &。B. 重载的运算符必须定义为类的成员函数。C. 运算符函数的第一个参数的类型是ostream &。D. 运算符函数有两个参数。10.假定AB为一个类，px为指向该类的一个含有该类析构函数的次数为()A. 0B. 1B. 0 和 1C. 1 和 0 D. 1 和 1C. n)o)。n个对象的动态数组的指针，则执行delete px;语句时共调用D. n+1二、填空题1-当使用保留字作为函数类型时，该函数不返回任何值。2. 执行char*p=new char( x*)操作后，p所指向的数据对象的值为。3. 当在程序中执行到语句时，将结束木次循环，执行下一次循环语句。4. 一个指针类型的对象

4、占用内存的个字节的存储空间。5. 假定用户为类AB定义了一个构造函数AB(intaa) (a=aa;,则系统(会/不会)为该类自动定义一个无参构造函数AB() %6. 已知语句coutsr的输出是hello,world ”,则执行语句 coutvs+%;”的输出结果为-7. 如果一个派生类的基类不止一个.则这种继承称为。8. 重载一个函数的条件是：该函数必须在参数的个数或参数对应的上与其它同名函数不相同。9. 假定用户为类AB定义了一个构造函数AB(intaa)(a=aa;)”，则系统(会/不会)为该类自动定义一个无参构造函数AB() %10 .假定用户为类AB定义了一个构造函数AB(inta

5、a=O):a(aa)0,则定义该类的对象时，可以有种不同的定义格式。三、程序填充题，根据题意在横线上填写合适的内容。1. 已知一个类的定义如下：#includeclass AA (int a10);int n;public:void SetA(int aa, int nn); / 初始化函数int MaxA();从数组a中前n个元素中查找最大值void SortA();”采用选择排序的方法对数组a中前n个元素进行从小到大排序；该类中MaxA()函数的实现如下，清在标号位置补充适当的内容。inKDint x=a0);for(int i=1; ix)-(2)-；T3)；(1)(2)2.类A的定义

6、class A ( int *a; int n; int MaxLen;public:A(): a(0), n(0), MaxLen(O) ( A(int *aa, int nn, int MM)( n=nn;MaxLen=MM; if(nMaxLen) exit(1); a=new intMaxLen; forfint i=0; in; i+) A(X(2) ；(1)(2)(3)无参构造函数带参构造函数；3.在下面一段类定义中，Derived类公有继承了基类Base。需要填充的函数由注释内容给出了功能。class Baseprivate:int mem1,mem2;“基类的数据成员publi

7、c:Base(int m1,int m2) (mem1=m1; mem2=m2;void output()(coutmem1 ,mem2；class Derived: public Base(private:int mem3;派生类木身的数据成员public:”构造函数，由ml和m2分别初始化mem 1和mem2.由m3初始化mem3 Derived(int m1,int m2, int m3);输出mem1,mem2和mem3数据成员的值void output()(1: coutmem3endl;；Derived:Derived(int m1,int m2, int m3):(2)( 3);

8、(2)(3)四、理解问答题，写出程序运行结果或程序(或函数)所能实现的功能。1. #includedouble f1(int n)(double sign=1,s=1;for(int i=2;i=n; i+) ( s+=sign/(i*i); sign*1;)return s;void main()(int a;cina;coutf1(a)endl;函数功能：2. char* f8(char* str1, const char* str2)(int i=0,j=0;while(str1 i) i+;while(str2j) str1 i+=str2j+;str1(i=,O,;return st

9、r1;)函数功能：3. #includevoid f2(int& x, int& y)(int z=x; x=y; y=z;)void f3(int* x, int* y)(int z=*x; *x=*y; *y=z;void main()(int x=10,y=26;coutx,y=x, yendl;即,y);coutx,y=x, yendl;f3(&x,&y);coutx,y=,x, yendl;x+; y-;丘(y，x)；coutx,y=x, yendl;运行结果：五、程序改借，清根据程序段或函数模块的功能改写个别地方的错误。下面是分数类fract的定义及测试主程序，在类定义及其友元函数

10、定义中有两处错误，更正错误后程序应显示41/28 ，请指出错误所在行的行号并给出改正意见。class fract(int den;分子int num; * 分母public:fract(int d=O,int n=1):den(d),num(n)(/1 行friend fract &operator+=(fract,fract&);112 行void show()( coutden7num;/3 行；/4行friend fract &operator+=(fract f1,fract f2)/5 行(/7行f1 .den=f1 .den*f2.num+f1 .num*f2.den;/8 行f1

11、.num,=f2.num;9行return f1;10 行void main()( fract fr(3,4); fr+=fract(5,7); fr.show();错误行的行号为和。(2分)分别改正为(4分)和(4分)参考答案：一、单选题1. B2. B3.A4.C5. D 6. D 7. A 8.A9. B 10. C二、填空题1. void 2. x 3.contiune 4. 45.不会6. pie 7.继承(或多重继承)8.类型9.不会10. 2三、程序填充题，根据题意在横线上填写合适的内容。评分标准：每空4分1. (1) AA:MaxA() (2) x=ai (3) return

12、x2. (1) a(i)=aai delete ga3. (1) Base:output() (2) Base(m1,m2)(3) mem3=m3四、理解问答题，写行结果或程序(或函数)所能实现的功能。1. 计算并输出1+,上*的值，其中a的值由键盘输入。2. 实现strcat函数0鱼血，把str2所指字符串连接到str1所指字符串的后面，并返回str1指针。3. x,y=10,26x,y=26,10x,y=10, 26x,y=25,11五、程序改销，清根据程序段或函数模块的功能改写个别地方的错误。25friend fract &operator+=(fract&.fract);fract &

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