404 lines
7.2 KiB
Markdown
404 lines
7.2 KiB
Markdown
# 目录
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- [目录](#目录)
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- [数组](#数组)
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- [一维数组](#一维数组)
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- [练习部分](#练习部分)
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- [斐波那契数列前10项](#斐波那契数列前10项)
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- [数据排序](#数据排序)
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- [冒泡](#冒泡)
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- [选择法](#选择法)
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- [进制转换](#进制转换)
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- [删除法求质数](#删除法求质数)
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- [二维数组](#二维数组)
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- [练习部分](#练习部分-1)
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- [行列互换](#行列互换)
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- [求最大值及其所在位置](#求最大值及其所在位置)
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- [求各行与各列的和](#求各行与各列的和)
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- [矩阵乘积](#矩阵乘积)
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- [字符数组](#字符数组)
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- [常用函数](#常用函数)
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- [练习部分](#练习部分-2)
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- [单词计数](#单词计数)
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- [多维数组](#多维数组)
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# 数组
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构造类型之一,连续存放。
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> [!warning]
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>
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> 时间关系,大量的练习题,没有做笔记,只记录了题目。自己思考加看课程足矣。
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## 一维数组
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1. 定义
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【存储类型】 数据类型 标识符 【下标】
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2. 初始化
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- 不初始化
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- 全部初始化
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- 部分初始化
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- `static`
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3. 元素引用
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数组名【下标】
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4. 数组名
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数组名是表示地址的**常量**,也是数组的起始位置。
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5. 数组越界
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### 练习部分
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#### 斐波那契数列前10项
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```c
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static void fibonacci(void)
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{
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int fib[10] = {1, 1};
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for (int i = 2; i < sizeof(fib) / sizeof(fib[0]); i++)
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{
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fib[i] = fib[i - 2] + fib[i - 1];
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}
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for (int i = 0; i < sizeof(fib) / sizeof(fib[0]); i++)
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{
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printf("fib[%d] = %d\n", i, fib[i]);
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}
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printf("\n");
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int i = 0;
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int j = sizeof(fib) / sizeof(fib[0]) - 1;
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int tmp;
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while (i < j)
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{
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tmp = fib[i];
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fib[i] = fib[j];
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fib[j] = tmp;
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i++;
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j--;
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}
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for (int i = 0; i < sizeof(fib) / sizeof(fib[0]); i++)
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{
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printf("fib[%d] = %d\n", i, fib[i]);
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}
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}
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```
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#### 数据排序
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##### 冒泡
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```c
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#define N 10
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static void sort1(void)
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{
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int a[N] = {12, 8, 45, 30, 98, 67, 2, 7, 68, 11};
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int tmp;
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for (int i = 0; i < N; i++)
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{
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printf("%d ", a[i]);
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}
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printf("\n");
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for (int i = 0; i < (N - 1); i++)
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{
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for (int j = 0; j < N - 1 - i; j++)
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{
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if (a[j] > a[j + 1])
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{
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tmp = a[j];
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a[j] = a[j + 1];
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a[j + 1] = tmp;
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}
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}
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}
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for (int i = 0; i < N; i++)
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{
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printf("%d ", a[i]);
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}
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printf("\n");
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}
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```
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##### 选择法
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```c
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static void sort2(void)
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{
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int a[N] = {23, 45, 90, 76, 13, 55, 76, 45, 3, 8};
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int i, j, k, tmp;
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for (int i = 0; i < N; i++)
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{
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printf("%d ", a[i]);
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}
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printf("\n");
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for (i = 0; i < N - 1; i++)
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{
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k = i;
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for (j = i + 1; j < N; j++)
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{
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if (a[j] < a[k])
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k = j;
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}
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if (i != k)
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{
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tmp = a[i];
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a[i] = a[k];
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a[k] = tmp;
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}
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}
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for (int i = 0; i < N; i++)
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{
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printf("%d ", a[i]);
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}
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printf("\n");
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}
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```
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#### 进制转换
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```c
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static void base_convert(void)
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{
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int num, base;
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int n[128];
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int i = 0;
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printf("Please enter the converted num:\n");
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scanf("%d", &num);
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printf("Please enter the base:\n");
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scanf("%d", &base);
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do
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{
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n[i] = num % base;
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num = num / base;
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i++;
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} while (num != 0);
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for (i--; i >= 0; i--)
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{
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if (n[i] >= 10)
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printf("%c", n[i] - 10 + 'A');
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else
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printf("%d", n[i]);
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}
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printf("\n");
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}
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```
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#### 删除法求质数
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```c
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static void primer(void)
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{
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char primer[1001] = {0};
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int i, j;
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for (i = 2; i < 1001; i++)
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{
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for (j = i * 2; j < 1001; j += i)
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primer[j] = -1;
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}
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for (i = 2; i < 1001; i++)
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{
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if (primer[i] == 0)
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printf("%d is a primer.\n", i);
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}
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}
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```
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## 二维数组
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1. 定义,初始化
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【存储类型】 数据类型 标识符 【行下标】 【列下标】
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> [!note]
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>
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> 初始化时,行号可省,列号不可省
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2. 元素引用
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数组名 【行标】【列标】
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3. 存储形式
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顺序存储,按行存储
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4. 对于二维数组的深入理解
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```c
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int a[M][N] = {1, 2, 3, 4, 5};
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int i, j;
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printf("a = %p\n", a);
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printf("a+1 = %p\n", a + 1);
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for (i = 0; i < M; i++)
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{
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for (j = 0; j < N; j++)
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{
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printf("%p --> %d\n", &a[i][j], a[i][j]);
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}
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printf("\n");
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}
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```
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输出:
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```c
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a = 0x7ffcc2376f80
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a+1 = 0x7ffcc2376f8c
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0x7ffcc2376f80 --> 1
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0x7ffcc2376f84 --> 2
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0x7ffcc2376f88 --> 3
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0x7ffcc2376f8c --> 4
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0x7ffcc2376f90 --> 5
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0x7ffcc2376f94 --> 0
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```
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`a+1`跨越了行而不是单个元素。
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### 练习部分
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#### 行列互换
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#### 求最大值及其所在位置
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#### 求各行与各列的和
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#### 矩阵乘积
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## 字符数组
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1. 定义,初始化,存储特点
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【存储类型】 数据类型 标识符 【下标】...
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- 单个字符初始化
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- 用字符串常量初始化
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2. 输入输出
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3. 常用函数
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### 常用函数
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1. `strlen`和`sizeof`
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```c
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size_t strlen(const char *s);
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// strlen函数可以获取字符串的长度,不包括尾'\0'
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// 以'\0'作为结束,所以对于"hello\0abc",abc就不计入
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// sizeof是完整的识别。
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char str[] = "hello\0abc";
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printf("%d\n", strlen(str));
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// out: 5
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printf("%d\n", sizeof(str));
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// out: 6
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```
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2. `strcpy`与`strncpy`
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```c
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char *strcpy(char *dest, const char *src);
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// 把src内容拷贝到dest,以'\0'结束,返回dest的地址
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// 当src大于dest,造成越界
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char *strncpy(char *dest, const char *src, size_t n);
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// 把src内容拷贝n个字节到dest
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strcpy(str, "abcde");
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puts(str);
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// out: abcde
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strncpy(str, "abcde", STR_SIZE);
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puts(str);
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// out: abcde
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```
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3. `strcat`与`strncat`
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```c
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char *strcat(char *dest, const char *src);
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// 把src补到dest后面
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// 需要保证dest空间足够
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char *strncat(char *dest, const char *src, size_t n);
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// 把src拿最多n个字符到dest
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strcat(str1, " ");
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strcat(str1, "world");
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// out:hello world
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strncat(str, " ", STR_SIZE);
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strncat(str, "world", STR_SIZE);
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// out:hello world
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```
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4. `strcmp`与`strncmp`
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```c
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int strcmp(const char *s1, const char *s2);
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// 返回两个字符串,ascii码的差值
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int strncmp(const char (s1, const char *s2, size_t n);
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// 指定比较前5个字符
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```
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### 练习部分
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#### 单词计数
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## 多维数组
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```c
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int a[2][3][4];
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```
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