Question

我一直在试图弄清楚如何到达以下程序的输出。我基本上想知道每次迭代后字符数组的状态。会有很大帮助。

#include<iostream.h>
#include<conio.h>
#include<ctype.h>
void main()
{
 char *name="ThE bESt meN wIN";
 for(int i=0;name[i]!='\0';i++)
 {
  if(islower(name[i]))
     name[i]=toupper(name[i]);
  else
  if(isupper(name[i]))
  {
   if(i%2==0)
    name[i]--;
   else
    name[i]=tolower(name[i-1]);
  }
  }
  cout<<name;
 getch();
}

输出是： SHD BbRT MEe WHh

Answer 1

将main返回类型更改为int并修改此行，使其指向可修改的字符数组：

char name[]="ThE bESt meN wIN";

然后一步一步地进行，这里是16次迭代中每次迭代的输出。数字指的是实际的数组索引。这应该会有所帮助。

如果不使用历史

，也许这更容易处理

initial
ThE bESt meN wIN


0123456789012345
ShE bESt meN wIN  0: T -> S

0123456789012345
SHE bESt meN wIN  1: h -> H

0123456789012345
SHD bESt meN wIN  2: E -> D

0123456789012345
SHD bESt meN wIN  3: space

0123456789012345
SHD BESt meN wIN  4: b -> B

0123456789012345
SHD BbSt meN wIN  5: E -> b

0123456789012345
SHD BbRt meN wIN  6: S -> R

0123456789012345
SHD BbRT meN wIN  7: t -> T

0123456789012345
SHD BbRT meN wIN  8: space

0123456789012345
SHD BbRT MeN wIN  9: m -> M

0123456789012345
SHD BbRT MEN wIN  10: e -> E

0123456789012345
SHD BbRT MEe wIN  11: N -> e

0123456789012345
SHD BbRT MEe wIN  12: space

0123456789012345
SHD BbRT MEe WIN  13: w -> W

0123456789012345
SHD BbRT MEe WHN  14: I -> H

0123456789012345
SHD BbRT MEe WHh  15: N -> h

请解释以下指针操作的输出

1 个答案: