有沒有大佬跟我解釋一下這題

#include <stdio.h>
int main()
{
??? /* 定義需要計算的日期 */
??? int year = 2008;
??? int month = 8;
??? int day = 8;
??? /*
???? * 請使用swtich語句，if...else語句完成本題
???? * 如有想看小編思路的，可以點擊左側任務中的“不會了怎么辦”
???? * 小編還是希望大家獨立完成哦~
???? */
??? int date = 0;
??? switch (month)
??? {
??????? case 12:??? date += 30;
??????? case 11:??? date += 31;
??????? case 10:??? date += 30;
??????? case 9:??? date += 31;
??????? case 8:??? date += 31;
??????? case 7:??? date += 30;
??????? case 6:??? date += 31;
??????? case 5:??? date += 30;
??????? case 4:??? date += 31;
??????? case 3:??? if(year%400 != 0 && year%4 == 0){
??????????????????????? date += 29;
??????????????????? }else{
??????????????????????? date += 28;}
??????? case 2:??? date += 31;
??????? default:??? date += 8,printf("2008年8月8日是該年的第%d天",date);break;
??? }
?return 0;
}

慕神5149347

2019-03-04

源自：C語言入門 4-17

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2 回答

EricGo101
2019-05-05

switch()括號里應該寫month-1,八月只過了八天，不能加上八月，執行順序是switch(7)，接著執行 case 7語句，date開始累加，case里沒有break語句，所以會一直往下執行，if應該在case 2寫，把case 1寫上，累加了1~7月的天數，最后在default里加上day(即date += day; )，使用day可以提高代碼的復用性。最后貼上我自己的代碼：

#include?<stdio.h>
int?main()?
{?
????/*?定義需要計算的日期?*/
????int?year?=?2008;
????int?month?=?8;
????int?day?=?8;
????/*
?????*?請使用switch語句，if...else語句完成本題
?????*?如有想看小編思路的，可以點擊左側任務中的“不會了怎么辦”
?????*?小編還是希望大家獨立完成哦~
?????*/
????int?sum?=?0;
?????
????switch(month?-?1){
????????
????????case?12:
????????????sum?+=?31;
????????case?11:
????????????sum?+=?30;
????????case?10:
????????????sum?+=?31;
????????case?9:
????????????sum?+=?30;
????????case?8:
????????????sum?+=?31;
????????case?7:
????????????sum?+=?31;
????????case?6:
????????????sum?+=?30;
????????case?5:
????????????sum?+=?31;
????????case?4:
????????????sum?+=?30;
????????case?3:
????????????sum?+=?31;
????????case?2:
????????????if(year%4?==?0)
????????????????sum?+=?29;
????????????else
????????????????sum?+=?28;
????????case?1:
????????????sum?+=?31;
????????default:
????????????sum?+=?day;
????????????break;
????}?
????printf("2008年8月8日是該年的第%d天",sum);
????
????return?0;
}

你自己理解理解！