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? ? {
? ? ? ??
? ? ? ? flag=-flag;
? ? ? ? sum=sum-i*flag;
? ? }??
2022-03-18
源自:C語言入門 4-10
正在回答
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flag=-flag的時(shí)候,也就是第一次循環(huán)sum那欄是sum+i,也就是1,第二次循環(huán)的時(shí)候用來循環(huán)的是第一次變化過的flag,也就是這時(shí)候用來循環(huán)的是-flag,而在第二次循環(huán)它又一次進(jìn)行了*-1的操作,變成了正的flag,這時(shí)候sun那欄就成了sum-i*flag,也就是成了-2。然后進(jìn)入第三次循環(huán)了。所以他完成了一負(fù)一正的運(yùn)行
誰如璀璨 提問者
舉報(bào)
C語言入門視頻教程,帶你進(jìn)入編程世界的必修課-C語言
1 回答flag*=-1不是一直是負(fù)的嗎為啥會(huì)算出來算出來
4 回答sum+=i*flag; flag*=-1; 這個(gè)代碼是怎么運(yùn)算的,想不明白
2 回答求大佬解答,這樣算不出來是25。。
1 回答有沒有大佬解釋一下這個(gè)?我不太明白
1 回答 i++; sum+=i*flag; flag*=-1;怎么算出來sum等于-50,實(shí)在想不出來這個(gè)式子
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2022-03-21
flag=-flag的時(shí)候,也就是第一次循環(huán)sum那欄是sum+i,也就是1,第二次循環(huán)的時(shí)候用來循環(huán)的是第一次變化過的flag,也就是這時(shí)候用來循環(huán)的是-flag,而在第二次循環(huán)它又一次進(jìn)行了*-1的操作,變成了正的flag,這時(shí)候sun那欄就成了sum-i*flag,也就是成了-2。然后進(jìn)入第三次循環(huán)了。所以他完成了一負(fù)一正的運(yùn)行