{int k=0;char c='A'; do { switch (c++) 输出结果为多少?请解释下,本人新手,希望不吝赐教。
答案:4 悬赏:50
解决时间 2021-01-26 01:19
- 提问者网友:鐵馬踏冰河
- 2021-01-25 22:06
{int k=0;char c='A'; do { switch (c++) 输出结果为多少?请解释下,本人新手,希望不吝赐教。
最佳答案
- 二级知识专家网友:白昼之月
- 2021-01-25 22:13
输出结果:4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4
全部回答
- 1楼网友:詩光轨車
- 2021-01-25 23:47
c='A',c++,CASE 'B'.K--,K=K+2,K=1;
C='B',C++,CASE C,K=K+2;K=3;
C='C',C++,CASE D,K=K%2,K=1;
C='D',C++,CASE E,K=K*10;K=10
C='E',C++,DEFAULT K=K/3;K=3;
C='G' printf("%d\n",k);
最后输出3
C='B',C++,CASE C,K=K+2;K=3;
C='C',C++,CASE D,K=K%2,K=1;
C='D',C++,CASE E,K=K*10;K=10
C='E',C++,DEFAULT K=K/3;K=3;
C='G' printf("%d\n",k);
最后输出3
- 2楼网友:天凉才是好个秋
- 2021-01-25 22:44
引用Y_雪莲的回答:
输出结果:4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4最后k=20,c=F,k=21,计算Fdefaultk=21\3=7,c=G,7+1=8
输出结果:4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4最后k=20,c=F,k=21,计算Fdefaultk=21\3=7,c=G,7+1=8
- 3楼网友:由着我着迷
- 2021-01-25 22:30
引用Y_雪莲的回答:
输出结果:4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4c++=E,k++,k=1 不等于2 cintinue是在判断
所以k=k*10=10
k++=11
输出结果:4
首先 c=‘A’,k=0: => k++,k=1 =>break;
=> c++ => c=B => k++,k=2
然后 c=B, k=2: => k--,k=1 => k+=2,k=3 => break;
=> c++ => c=C => k++,k=4
然后 c=C, k=4: => k+=2,k=6 => break;
=> c++ => c=D => k++,k=7
然后 c=D, k=7: => k=k%2,k=1 => continue;
=> c++ => c=E => k++,k=2
然后 c=E, k=2: => k=k*10,k=20 => break;
=> c++ => c=F =>k++,k=21
然后 c=F, k=11:=> k=k/3,k=3
=> c++ => c=G => k++,k=4
因为 c=G 不满足 while (c<'G') 循环条件,所以循环结束
输出 4c++=E,k++,k=1 不等于2 cintinue是在判断
所以k=k*10=10
k++=11
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