#include <stdio.h>
void inverte(int *x, int n)
{
int *p, temp, *i, *j, m = (n - 1) / 2;
i = x;
j = x + (n - 1);
p = x + m;
for (i = x, j = x + (n - 1); i <= p; i++, j--)
{
temp = *i;
*i = *j;
*j = temp;
}
return 0;
}
int main()
{
void inverte(int *x, int n);
int i;
int a[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
printf("原数组为：\n");
for (i = 0; i < 10; i++)
{
printf("%d ", a[i]);
}
printf("\n");
inverte(a, 10);
printf("交换后的数组为：\n");
for (i = 0; i < 10; i++)
{
printf("%d ", a[i]);
}
printf("\n");
return 0;
}
方法二：
#include <stdio.h>
int main()
{
void inv(int b[], int n);
int a[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int i;
printf("原数组为：\n");
for (i = 0; i < 10; i++)
{
printf("%d ", a[i]);
}
printf("\n");
inv (a, 10);
printf("交换后的数组为：\n");
for (i = 0; i < 10; i++)
{
printf("%d ", a[i]);
}
printf("\n");
return 0;
}
void inv (int b[], int n)
{
int i, j, temp, m;
m = (n - 1) / 2;
for (i = 0; i <= m; i++)
{
j = n - 1 - i;
temp = b[i];
b[i] = b[j];
b[j] = temp;
}
}
