Question

下面的这个程序，可以用c++中的T 函数实现,可以减少一个函数,谢谢

#include<stdio.h>
#define M 7
#define N 7
int *small(int *x,int *y);
char *big(char *m,char *n);
void main()
{
int str1[M]={9,5,6,7,4,10,8};
char str2[N]={'c','i','h','m','t','a','g'};
int i,j;
int *p1,*p2;
char *p3,*p4;
p1=str1;
p2=&str1[6];
p3=str2;
p4=&str2[6];
small(p1,p2);
big(p3,p4);
for(i=0;i<N;i++)
printf("%3ld",str1[i]);
printf("\n");
for(j=0;j<M;j++)
printf("%c",str2[j]);
printf("\n");
}
int *small(int *x,int *y)
{
int k,i;
for( i=0;i<M/2;i++,x++,y--)
{
k=*x;
*x=*y;
*y=k;
}
return 0;
}
char *big(char *m,char *n)
{
int t,j;
for( j=0;j<N/2;j++,m++,n--)
{
t=*m;
*m=*n;
*n=t;
}
return 0;
}

Answer 1

#include<stdio.h>
#define M 7
#define N 7
template <class T>
T *change(T *x,T *y)
{
int k,i;
for( i=0;i<M/2;i++,x++,y--)
{
k=*x;
*x=*y;
*y=k;
}
return 0;
}
void main()
{
int str1[M]={9,5,6,7,4,10,8};
char str2[N]={'c','i','h','m','t','a','g'};
int i,j;
int *p1,*p2;
char *p3,*p4;
p1=str1;
p2=&str1[6];
p3=str2;
p4=&str2[6];
//small(p1,p2);
//big(p3,p4);
change<int>(p1,p2);
change<char>(p3,p4);
for(i=0;i<N;i++)
printf("%3ld",str1[i]);
printf("\n");
for(j=0;j<M;j++)
printf("%c",str2[j]);
printf("\n");
}