c程序设计语言作业题及解析

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1、吉+A-去+么+(-1)f叫二"+言第三+@^2013年4月考试C++程序设计语次作业丄++丄x+1x2-2x3+3x4-4+填空题(本大题共20分’共5小题，每小题4分)1.C++程序由函数构成，它总是由主函数开始.2.假定要访M—个结构指针p所指对象中的b指针成员所指的对象，则表示方法为.3.己知WCZ1的ASCII码为65(90,当执行"charch=14*5+2;cout«ch<4.以下函数完成求表达式宿命sum(x)=-的值，请填空使之完成此功能：floatsunffloatx){floats=0.0;intsign=l;for(in

2、tj=l；；i++){t=t*x;s=s+[blank];sign=-sign;}5.以下函数完成求两个数nl和n2的最大公约数.#includeintfac(intnl,intn2){inttmp;iff){tmp=nl;nl—n2•zn2—tmp參/}while(){tmp—nl%n2;nl=n2參/n2=tmp;}returnn2;}程序阅读题(本大题共30分,共3小题，每小题10分)1.写出下面程序运行结果。include〈iostream.h>include#definePI3.1415926classT{chardo

3、ublepublic:T(doublea,char{x=a;flag=f;}char{return}double{doublereturn}double{doublereturn}};void{chardoubleTdoubleif(t.getflag()=='s')p=T::Sin;elsep=T::Cos;cout«"resultis:}2.指出下面程序段功能。flag;x；f)getflag()flag;Sin()y=x*PI/180;sin(y);Cos()y=x*PI/180;cos(y);main()c='c';r=60;t(r,c

4、);(T：：*p)()；if(t.getflag()==,c,)"«(t.*p)()«endl;#includeh>//include//definePI3.1415926classsincos{doublex;public:sincos(doublea){x=a;}doublecomput(double{x=x*PI/180;z=sin(x);returncos(x);}};3.写出typedef#define#include//includeclass{Tpublic:ArrType(T{a=x;}T{if(i<0

5、

6、i〉=N){cout«

7、"数exit(l);}return}}；void{TF面程序运行intN&operator[](int组越界错误T；5ArrTypea;i)"«endl;a[U；main()a[N】={l,2,3A5};arr(a);i=O;i

8、任何异常吗？下而的facfintnlfac,写nl,ml=nl,<出该函数的intn2n2)m2=n2;tmp;)nl;n2tmp}while){tmpnln2}returnnl%n2nln2n2;tmp;}3.4述streambuf序，写出程include#流类序运行include简程-•Gpa果sh用结作出ssla输d的时public:{shape(doublex,doubley):xCoord(x),yCoord(y)}virtual{return}protected:doubleArea()const0.0;doublexCoord,y

9、Coord;｝；classAA:publicshape{public:AAfdoublex,doubley.doubler):shape(x,y)zrad(r){}virtual{return}protected:doubleArea()const3.0radrad;doublerad;};classBB:publicshape{public:BB(doublexl,doubleyl,doublex2?doubley2):shape(xl,yl),x2Coord(x2)?y2Coord(y2){}virtualdoubledoubleArea(

10、)const;protected:BB:Area()doublex2Coord,y2Coord;};constreturnfabs((xCoord-x