通过调用函数来交换x y 的值，用指针来完成！谢谢高手帮忙！
#include<iostream.h>
void main()
{int *a,*b;
int x,y;
int swap(int *m,int* n);
cin>>x;
cin>>y;
a=&x;
b=&y;
swap(a,b);
cout<<*a;
cout<<*b;
}
int swap(int *m,int *n)
{
int *t;
t=m;
m=n;
n=t;
// cout<<*m;
// cout<<*n;
return(*m,*n);
}

int *t;

-->

int t;

不可以呀！请高手再给看看呀！谢谢了！

类型严重不匹配，指针与int怎么可以赋值？？？



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…………这个……是C++?

建议楼主先去把C++ Primer看3至5遍。

[此贴子已经被作者于2007-9-18 21:26:09编辑过]

新手～不懂

#include <iostream>
using namespace std;

int main()
{
int* a, * b;
int x, y, z;
int swap(int* m, int* n);
cin >> x;
cin >> y;
a = &x;
b = &y;
z = swap(a, b);
cout << * a << endl;
cout << * b << endl;

cout<<z<<endl;

return 0;
}

int swap(int* m, int* n)
{
int t;
t = * m;
*m = * n;
*n = t;
// cout<<*m;
// cout<<*n;
return (*m, * n); // returns value of *m. Is this what you want?
}

/**
Hi sister 凌冰月:

read more about passing by value and passing by reference/address.

Output:
5 6
6
5
5
Press any key to continue . . .
*/

C++是可以兼容C的.

#include<iostream.h>
void swap(int *rx,int *ry);
int main()
{
int x,y;
cout<<"please enter two number:";
cin>>x;
cin>>y;
cout<<"before the swap...";
cout<<"x:"<<x<<"y:"<<y<<endl;
swap(x,y);
cout<<"after the swap...."
cout<<"x:"<<x<<"y:"<<y<<endl;
return 0;
}
void swap(int *rx,int *ry)
{
int temp;
temp=*rx;
*rx=*ry;
*ry=temp;
}

#include<iostream.h>
void main()
{int *a,*b;
int x,y;
int swap(int *m,int* n);
cin>>x;
cin>>y;
a=&x;
b=&y;
swap(a,b);
cout<<*a;
cout<<*b;
}
int swap(int *m,int *n)
{
int t;
t=*m;
*m=*n;
*n=t;
// cout<<*m;
// cout<<*n;
return(*m,*n);
}