#include <stdio.h>int main()
{//打印數組中的最大值int arr[10];int max,i;for (i = 0; i < 10; i++){scanf_s("%d", &arr[i]);}max = arr[0];for (i = 0; i < 10; i++){if(max < arr[i + 1]){max = arr[i + 1];}}printf("數組中最大值：%d\n", max);return 0;
}

#include <stdio.h>int main()
{//交換數組中的數據逆數組int arr[10];int i, t;int n = sizeof(arr) / sizeof(int);printf("請輸入10個數據\n");for (i = 0; i < n; i++){scanf_s("%d", &arr[i]);}for (i = 0; i < n / 2; i++){t = arr[i];arr[i] = arr[n - 1 - i];arr[n - 1 - i] = t;}for (i = 0; i < n; i++){printf("%d", arr[i]);}return 0;
}

#include <stdio.h>int main()
{//數組中的數據排大小int arr[10];int i, t, j ;t = 0;int n = sizeof(arr) / sizeof(int);printf("請輸入10個數據\n");for (i = 0; i < n; i++){scanf_s("%d", &arr[i]);}for (i = 0; i < n-1; i++){for (j = 0; j < n - 1 - i; j++){if(arr[j]>arr[j+1]){t = arr[j];arr[j] = arr[j + 1];arr[j + 1] = t;}}}for (i = 0; i < n; i++){printf("%d", arr[i]);}return 0;
}

#include <stdio.h>
#include<stdbool.h>
int main()
{//打印數組中的最大值int arr[3][3] = { {3,6,29},{2,6,1},{3,22,44} };int i, j;int max = arr[0][0];int line, column;for (i = 0; i < 3; i++){for (j = 0; j < 3; j++){if (max<arr[i][j]){max = arr[i][j];line = i;column = j;}printf("%2d ", arr[i][j]);}}printf("max= %d %d %d\n", max, line, column);return 0;
}

#include <stdio.h>
#include<stdbool.h>
int main()
{//指針的數組逆運算int a[] = {1,5,8,10,20,30};int i, n;int* p, * q;int t;n = sizeof(a) / sizeof(int);p = a;q = &a[n - 1];for (i = 0; i < n / 2; i++){t = *p;*p = *q;*q = t;p++;q--;}for (i = 0; i < n; i++){printf("%d ", a[i]);}printf("\n");return 0;
}

?

#include <stdio.h>
#include<stdbool.h>
int main()
{//利用指針訪問地址和元素int a[3][4] = { {1,2,3,4},{4,5,6,7},{6,7,8,9} };int i, j;int n = sizeof(a) / sizeof(a[0]);//計算行數int m = sizeof(a[0]) / sizeof(int);//每行的元素個數int* p;for (i = 0; i < n; i++){for (j = 0; j < m; j++){printf("%p", &a[i][j]);//地址}printf("\n");}p = a;for (i = 0; i < n * m; i++){printf("%d\n", *(p + i));}return 0;
}

#include <stdio.h>
#include<stdbool.h>
int main()
{//利用指針訪問地址和元素int a[3][4] = { {1,2,3,4},{4,5,6,7},{6,7,8,9} };int (*p)[4] = a;int i, j;printf("%p %p\n", a, a + 1);printf("%p %p\n", a[0], a[0 + 1]);int n = sizeof(a) / sizeof(a[0]);//計算行數int m = sizeof(a[0]) / sizeof(int);//每行的元素個數for (i = 0; i < n; i++){for (j = 0; j < m; j++){printf("%d %d %d",p[i][j],*(p[i]+j),*(*(p+i)+j));}printf("\n");
}return 0;
}