题目是：找一个二维数组中的鞍点，即该位置上的元素在该行上最大，在该列上最小。也可能没有鞍点
#include<stdio.h>
void main()
{
        void max(int *(p)[4],int n);
        int a[3][4]={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
        int s,t1,t2,t3;
        t1=max(a[0],0);
        t2=max(a[1],1);
        t3=max(a[2],2);
        s=t1+t2+t3;
        if(s==3)
        printf("no");
}
void max(int *(p)[4],int n)
{int i,t,s1,s2,q,g;
t=0;
    for(i=0;i<4;i++)
    {       if(*(*(p)+i)>t)
            {t=*(*(p)+i);
            q=i;
            }
        if(n=0)s1=1,s2=2;
       else if(n=1)s1=0,s2=2;
        else(n=2)s1=0,s2=1;
        if(*(*(p)+q)<*(*(p+s1)+q)&&*(*(p)+q)<*(*(p+s2)+q))
        printf("%d",*(*(p)+q);
        else
        g=g+1;
        retarn(g);
}

#include<stdio.h>
void main()
{
        void max(int *(p)[4],int n); /*你定义的函数有反回值，这又申明void，改int,参数不对，改(*p)[4]*/
        int a[3][4]={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
        int s,t1,t2,t3;
        t1=max(a[0],0);
        t2=max(a[1],1);
        t3=max(a[2],2);
        s=t1+t2+t3;
        if(s==3)
        printf("no");
}
void max(int *(p)[4],int n)  /*这里void改int,参数不对，改(*p)[4]*/
{int i,t,s1,s2,q,g;
t=0;
    for(i=0;i<4;i++)
    {       if(*(*(p)+i)>t)
            {t=*(*(p)+i);
            q=i;
            }
    }                         /*少个花括号*/
        if(n=0)s1=1,s2=2;    /*if里表达式错，是＝＝*/
       else if(n=1)s1=0,s2=2;  /*if里表达式错，是＝＝*/
        else(n=2)s1=0,s2=1;  /*少了if    表达式也错，是＝＝*/
        if(*(*(p)+q)<*(*(p+s1)+q)&&*(*(p)+q)<*(*(p+s2)+q))
         printf("%d",*(*(p)+q);  /*少个'）'括号*/
        else
        g=g+1;  /*变g没初始值*/
        retarn(g);  /* 反回函数打错，应该是return */
}


[ 本帖最后由 UserYuH 于 2009-9-3 14:39 编辑 ]