调试问题zenmeban

Question

为什么把四个seek函数放在（p+2）之前都能显示，放在之后就不可以了.
#include <stdio.h>
int main(void)
{
double aver(int p, int n);
void seek(int* p, int n);
void good(int* p, int n);
int a[6] = { 56,78,45,50,89,1 };
int b[6] = { 89,86,90,98,95 ,2};
int c[6] = { 79,88,99,78,67 ,3};
int d[6] = { 95,78,90,99,89 ,4};
int* p;
p = a;
*(p + 1) = b[0];
*(p + 2) = c[0];
seek(a, 5); seek(b, 5); seek(c, 5); seek(d, 5);
(p + 3) = d[0];
seek(a, 5); seek(b, 5); seek(c, 5); seek(d, 5);
printf("%lf \n",aver(p, 4));
good(a, 5); good(b, 5); good(c, 5); good(d, 5);
return 0;
}
double aver(int p, int n)
{
int i,sum;
sum = 0;
double a;
for (i = 0; i < n; i++)
sum = sum + p[i];
a = (double)sum / n;
return a;
}
void seek(int* p, int n)
{
int i;
int a = 0;
for (i = 0; i < n; i++)
if (p[i] < 60)
a = a + 1;
if (a > 1)
{
for (i = 0; i < n; i++)
printf("%d \n",p[i]);
aver(p, n);
printf("%lf \n",aver(p,n));
printf("%d \n",p[5]);

}
return;

}
void good(int* p, int n)
{
int i, j = 0;
for (i = 0; i < n; i++)
if (p[i] > 85)
j++;
if (aver(p, n) > 90)
printf("%d ", p[5]);
else if (j == 5)
printf("%d ", p[5]);
else printf("NULL ");
return;
}

Answer 1

代码没什么问题啊，不显示什么？void seek(int* p, int n) 函数实现有不及格的才显示，2 3 4 都没有不及格的，当然不显示了。

#include <stdio.h>
int main(void)
{
    double aver(int* p, int n);
    void   seek(int* p, int n);
    void   good(int* p, int n);
    int a[6] = { 56,78,45,50,89,1 };
    int b[6] = { 89,86,90,98,95 ,2 };
    int c[6] = { 79,88,99,78,67 ,3 };
    int d[6] = { 95,78,90,99,89 ,4 };
    int* p;
    p = a;
    *(p + 1) = b[0];
    *(p + 2) = c[0];
    seek(a, 5); seek(b, 5); seek(c, 5); seek(d, 5);
    *(p + 3) = d[0];
    seek(a, 5); seek(b, 5); seek(c, 5); seek(d, 5);
    printf("%lf \n", aver(p, 4));
    good(a, 5); good(b, 5); good(c, 5); good(d, 5);
    return 0;
}
double aver(int* p, int n)
{
    int i, sum;
    sum = 0;
    double a;
    for (i = 0; i < n; i++)
        sum = sum + p[i];
    a = (double)sum / n;
    return a;
}
void seek(int* p, int n)
{
    int i;
    int a = 0;
    for (i = 0; i < n; i++)
        if (p[i] < 60)
            a = a + 1;
    if (a > 1)
    {
        for (i = 0; i < n; i++)
            printf("%d \n", p[i]);
        //aver(p, n);
        printf("%lf \n", aver(p, n));
        printf("%d \n", p[5]);
    }
    return;
}
void good(int* p, int n)
{
    int i, j = 0;
    for (i = 0; i < n; i++)
        if (p[i] > 85)
            j++;
    if (aver(p, n) > 90)
        printf("%d ", p[5]);
    else if (j == 5)
        printf("%d ", p[5]);
    else 
        printf("NULL ");
    return;
}