今日主要学习的是分支语句
- #define _CRT_SECURE_NO_WARNINGS 1
- //if else分支语句
- #include
- //代码1
- int main()
- {
- int age = 0;
- scanf("%d", &age);
- if (age < 18)
- {
- printf("未成年\n");
- }
- }
- //代码2
- #include
- int main()
- {
- int age = 0;
- scanf("%d", &age);
- if (age < 18)
- {
- printf("未成年\n");
- }
- else
- {
- printf("成年\n");
- }
- }
- //代码3
- #include
- int main()
- {
- int age = 0;
- scanf("%d", &age);
- if (age < 18)
- {
- printf("少年\n");
- }
- else if (age >= 18 && age < 30)
- {
- printf("青年\n");
- }
- else if (age >= 30 && age < 50)
- {
- printf("中年\n");
- }
- else if (age >= 50 && age < 80)
- {
- printf("老年\n");
- }
- else
- {
- printf("老寿星\n");
- }
- }
- #include
- int main()
- {
- int a = 0;
- int b = 2;
- if (a == 1)
- {
- if (b == 2)
- {
- printf("hehe\n");
- }
- }
- else
- {
- printf("haha\n");
- }
- return 0;
- }
- //else是和它离的最近的if匹配的
- /*练习
- 1. 判断一个数是否为奇数
- 2. 输出1 - 100之间的奇数*/
- #include
- int main()
- {
- int a = 1;
- while (a <= 60)
- {
- if (a % 2 == 1)
- printf("%d\n", a);
- a++;
- }
- return 0;
- }
- #include <stdio.h>
- int main()
- {
- int a = 1;
- while (a <= 60)
- {
- if (a % 2 == 1)
- printf("%d\n", a);
- a++;
- }
- return 0;
- }
- //switch语句
- #include <stdio.h>
- int main()
- {
- int day = 0;
- switch (day)
- {
- case 1:
- printf("星期一\n");
- break;//break语句 的实际效果是把语句列表划分为不同的分支部分。
- case 2:
- printf("星期二\n");
- break;
- case 3:
- printf("星期三\n");
- break;
- case 4:
- printf("星期四\n");
- break;
- case 5:
- printf("星期五\n");
- break;
- case 6:
- printf("星期六\n");
- break;
- case 7:
- printf("星期天\n");
- break;
- }
- return 0;
- }
- //需求变了:
- //1. 输入1 - 5,输出的是“weekday”;
- //2. 输入6 - 7,输出“weekend”
- #include <stdio.h>
- //switch代码演示
- int main()
- {
- int day = 0;
- switch (day)
- {
- case 1:
- case 2:
- case 3:
- case 4:
- case 5:
- printf("weekday\n");
- break;
- case 6:
- case 7:
- printf("weekend\n");
- break;//编程好习惯在每个 switch 语句中都放一条default子句是个好习惯,甚至可以在后边再加一个 break
- }
- return 0;
- }
- //练习
- #include <stdio.h>
- int main()
- {
- int n = 1;
- int m = 2;
- switch (n)
- {
- case 1:
- m++;
- case 2:
- n++;
- case 3:
- switch (n)
- {//switch允许嵌套使用
- case 1:
- n++;
- case 2:
- m++;
- n++;
- break;
- }
- case 4:
- m++;
- break;
- default:
- break;
- }
- printf("m = %d, n = %d\n", m, n);
- return 0;
- }
- //输出结果应是m=5,n=3