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width="85%" class="body18black">C++习题与解析(重载-04)</TD>
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<TD class="body12black" >题6.设计一个三角形类Triangle，包含三角形三条边长的私有数据成员，另有一个重载运算符“+”，以实现求两个三角形对象的面积之和。<BR>
解：<BR>
在Triangle类中设计一个友元函数operator+(Triangle t1,Triangle
t2)，它重载运算符"+"，返回t1和t2两个三角形的面积之和。<BR>本题程序如下：<BR>#include<iostream.h><BR>#include<math.h><BR>class
Triangle<BR>{<BR> int
x,y,z;<BR> double
area;<BR>
public:<BR>
Triangle(int i,int j,int
k)<BR>
{<BR>
double
s;<BR>
x=i;y=j;z=k;<BR>
s=(x+y+z)/2.0;<BR>
area=sqrt(s*(s-x)*(s-y)*(s-z));<BR>
}<BR>
void
disparea()<BR>
{<BR>
cout<<"Area="<<area<<endl;<BR>
}<BR>
friend double operator+(Triangle t1,Triangle
t2)<BR>
{<BR>
return
t1.area+t2.area;<BR>
}<BR>};<BR>void main()<BR>{<BR>
Triangle
t1(3,4,5),t2(4,5,6);<BR> double
s;<BR> cout<<"t1:";
t1.disparea();<BR>
cout<<"t2:";
t2.disparea();<BR>
s=t1+t2;<BR>
cout<<"总面积="<<s<<endl;<BR>}<BR>本程序执行结果如下：<BR>t1:Area=6<BR>t2:Area=9.92157<BR>总面积=15.9216<BR><BR>-----------------------------------------------------------<BR><BR><FONT
color=#0000ff>题7.习题6的重载运算符“+”友元函数只能返回两个三角形的面积之和，不能计算三个三角形的面积之和，改进一下，使之能计算任意多个三角形的面积之和。<BR></FONT>解：<BR>
习题6的重载运算符为什么不能计算3个三角形的面积之和呢？对于式子:s=t1+t2+t3，先计算t1+t2，返回一个double数然后再进行该double数+t3的计算，显然没有这样的重载运算符“+”友元函数，只需要添加这样重载运算符“+”友元函数即可。<BR>本题程序如下：<BR>#include<iostream.h><BR>#include<math.h><BR>class
Triangle<BR>{<BR> int
x,y,z;<BR> double
area;<BR>
public:<BR>
Triangle(int i,int j,int
k)<BR>
{<BR>
double
s;<BR>
x=i;y=j;z=k;<BR>
s=(x+y+z)/2.0;<BR>
area=sqrt(s*(s-x)*(s-y)*(s-z));<BR>
}<BR>
void
disparea()<BR>
{<BR>
cout<<"Area="<<area<<endl;<BR>
}<BR>
friend double operator+(Triangle t1,Triangle
t2)<BR>
{<BR>
return
t1.area+t2.area;<BR>
}<BR>
friend double operator+(double d,Triangle
t)<BR>
{<BR>
return
d+t.area;<BR>
}<BR>};<BR>void main()<BR>{<BR>
Triangle
t1(3,4,5),t2(4,5,6),t3(5,6,7),t4(6,7,8);<BR>
double s;<BR> cout<<"t1:";
t1.disparea();<BR>
cout<<"t2:";
t2.disparea();<BR>
cout<<"t3:";
t3.disparea();<BR>
cout<<"t4:";
t4.disparea();<BR>
s=t1+t2+t3+t4;<BR>
cout<<"总面积="<<s<<endl;<BR>}<BR>本程序的执行结果如下：<BR>t1:Area=6<BR>t2:Area=9.92157<BR>t3:Area=14.6969<BR>t4:Area=20.3332<BR>总面积=50.9517<BR><BR>-------------------------------------------------------<BR><BR><FONT
color=#0000ff>题8.设计一个学生类student,包括姓名和三门课程成绩，利用重载运算符”+“将所有学生的成绩相加放在一个对象中，再对该对象求各门课程的平均分。<BR></FONT>解：<BR>#include<iostream.h><BR>#include<iomanip.h><BR>#include<string.h><BR>class
student<BR>{<BR> char
name[10];<BR> int
deg1,deg2,deg3;<BR>
public:<BR>
student(){}<BR>
student(char na[],int d1,int d2,int
d3)<BR>
{<BR>
strcpy(name,na);<BR>
deg1=d1;deg2=d2;deg3=d3;<BR>
}<BR>
friend student operator+(student s1,student
s2)<BR>
{<BR>
static student
st;<BR>
st.deg1=s1.deg1+s2.deg1;<BR>
st.deg2=s1.deg2+s2.deg2;<BR>
st.deg3=s1.deg3+s2.deg3;<BR>
return
st;<BR>
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