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width="85%" class="body18black">C++习题与解析(引用-03)</TD>
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<TD class="body12black" >题1.分析以下程序的执行结果<BR>
#include<iostream.h><BR>void
swap(int &x,int
&y)<BR>{<BR> int
temp;<BR> temp=x;
x=y; y=temp;<BR>}<BR>void
main()<BR>{<BR> int
x=10,y=20;<BR> swap(x,y);<BR> cout<<"x="<<x<<",y="<<y<<endl;<BR>}<BR>解：<BR> 这里的函数采用引用调用的方式，所以输出为：x=20,y=10<BR>注意：在函数调用里，引用调用与传址调用的效果相同，但更加简洁直观。<BR><BR>----------------------------------------------------------<BR><BR><FONT
color=#0000ff>题2.分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>void
main()<BR>{<BR> int
a[]={10,20,30,40},*pa=a;<BR> int
*&pb=pa;<BR> pb++;<BR> cout<<*pa<<endl;<BR>}<BR>解：<BR> pa为数组的指针，首先指向a[0]，pb是pa的引用，当执行pb++时，也使pa指向了a[1]，所以输出为:20<BR><BR>-------------------------------------------------------------<BR><BR><FONT
color=#0000ff>题3.分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>class
Sample<BR>{<BR> int
x;<BR> public:<BR> Sample(){};<BR> Sample(int
a){x=a;}<BR> Sample(Sample
&a){x=a.x++
+10;}<BR> void
disp(){cout<<"x="<<x<<endl;}<BR>};<BR>void
main()<BR>{<BR> Sample
s1(2),s2(s1);<BR> s1.disp();<BR> s2.disp();<BR>}<BR>解：<BR>
Sample类的Sample(Sample
&a)构造函数是一个拷贝构造函数，将a对象的x增1然后加上10后赋给当前对象的x，由于a是引用对象，所以输出为：<BR>
x=3 //
++运算的结果<BR>
x=12 //
2+10<BR><BR>--------------------------------------------------------------<BR><BR><FONT
color=#0000ff>题4.分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>class
Sample<BR>{<BR> int
x,y;<BR>
public:<BR>
Sample(){x=y=0;}<BR>
Sample(int i,int
j){x=i;y=j;}<BR>
void copy(Sample
&s);<BR>
void setxy(int i,int
j){x=i;y=j;}<BR>
void
print(){cout<<"x="<<x<<",y="<<y<<endl;}<BR>};<BR>void
Sample::copy(Sample
&s)<BR>{<BR>
x=s.x;y=s.y;<BR>}<BR>void func(Sample s1,Sample
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