為什么這么寫
int i=0;
while(i<10)
{
????i++
????prinft("第%d遍書寫:computer\n",i);
i++不是先賦值再運算嗎,那第一次運算后的結(jié)果難道是:第0遍書寫computer?第0遍怎么感覺怪怪的?此時i=1對吧
int i=0;
while(i<10)
{
????i++
????prinft("第%d遍書寫:computer\n",i);
i++不是先賦值再運算嗎,那第一次運算后的結(jié)果難道是:第0遍書寫computer?第0遍怎么感覺怪怪的?此時i=1對吧
2021-07-11
舉報
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2021-07-26
為0的i++到printf這步就已經(jīng)為1了,到i為9的時候printf已經(jīng)為10,所以剛好10遍。
2021-07-23
#include <stdio.h>
int main()
{
? ? int i=0;
? ? while(i<10)
? ? {
? ? ? ? printf("第%d遍書寫:computer\n",++i);
? ? }
? ? return 0;
}
2021-07-12
#include <stdio.h>
int main()
{
? ? int i,sum=0;
? ? i=1;
? ? while(i<=100)? //循環(huán)條件應(yīng)該是什么呢?
? ? {
? ? ? ? sum=sum+i;?
? ? ? ? i++;//這里是不是應(yīng)該改變循環(huán)變量的值
? ? }
? ? printf("100以內(nèi)所有整數(shù)之和為:%d\n", sum);
? ? return 0;
}
首先,你的循環(huán)條件寫錯了,應(yīng)為while(i<=100)說明i的值僅在100以內(nèi)取,接下來是循環(huán)體,當(dāng)i滿足循環(huán)調(diào)節(jié)就會執(zhí)行sum=i+sum的式子,執(zhí)行完后會有i++(注意:i++只能放在那個式子后面,代表執(zhí)行完式子之后執(zhí)行i++),因為int i=1? ?int sum=0,那么第一次循環(huán)就是sum=1+0=1,然后i++使i自增1,即為2,使第二次循環(huán)中的i值就為第一次循環(huán)結(jié)束的i++的值,為2,且第二次循環(huán)的sum值為第一次循環(huán)結(jié)束后的sum值,即為1,那么第二次循環(huán)的結(jié)果是sum,1+2,然后后面的循環(huán)如上,就形成了1+2+3+...+100的結(jié)果,為5050