教材没对这个做任何解释直接拿来用了~看不懂！
#include<stdio.h>
void main()
{
    float score[][4]={{60,70,80,90},{59,89,67,88},{34,78,90,66}};
    float*search(float(*pointer)[4],int n);
    float *p;
    int i,m;
    printf("Enter the number of student:");
    scanf("%d",&m);
    printf("the score of No.%d are:\n",m);
    p=search(score,m);
    for(i=0;i<4;i++)
    printf("%5.2f\t",*(p+i));
    printf("\n");
}
float *search(float(*pointer)[4],int n)
{
float *pt;
pt=*(pointer+n); ？？此处的*（pointer+n）代表score[n]，而score是二维数组，所以score[n]表示第n行的首地址，
                                            所以这里*（pointer+n）分别是各行的首地址
return pt;
}
请问这里的指针int (*p)[n]是神马意思~迷糊了~括号里面的n代表的是神马?如果把这个二维数组当成一维来看，括号里面也应该是3才对撒

我对程序作了修改~改成这样我懂~
#include<stdio.h>
void main()
{
    float score[][4]={{60,70,80,90},{59,89,67,88},{34,78,90,66}};
    float*search(float*pointer,int n);
    float *p;
    int i,m;
    printf("Enter the number of student:");
    scanf("%d",&m);
    printf("the score of No.%d are:\n",m);
    p=search(score[0],m);
    for(i=0;i<4;i++)
    printf("%5.2f\t",*(p+i));
    printf("\n");
}
float *search(float*pointer,int n)
{
float *pt;
pt=pointer+4*n;
return pt;
}

完善程序：显示编号No.的成绩并求出不及格人数的编号：
#include<stdio.h>
int j;
void main()
{
    int i,n,*p;
    int x[3][4]={60,70,80,90,56,89,67,88,34,78,90,66};
    int *fun(int*x,int y);
    int *search(int*x);
    printf("Input the number No.");
    scanf("%d",&n);
    p=fun(x[0],n-1);
    printf("The grades is:");
    for(i=0;i<4;i++)
    printf("%d\t",*(p+i));
    printf("\n");
    for(j=0;j<3;j++)
    {
    p=search(x[j]);
    for(i=0;i<4;i++)
    {
    if(*(p+i)<60)
    {
    printf("The No.%d is a failure!\n",j+1);
    break;
    }
    }
    }
}

int *fun(int*x,int y)
{
    int *z;
    z=x+4*y;
    return z;
}

int *search(int*x)
{
return (x+4*j);
}

木人~~~~~~(>_<)~~~~

懂了~谢谢！