在CS50课程2020的pset2中检查代码中的替换问题时,我一直在处理键中的重复字符时出错。我的代码和更多细节如下--有人能帮我吗?谢谢
它给我的错误消息是
:( handles duplicate characters in key
timed out while waiting for program to exit
当我检查代码中是否有重复字符时,它似乎工作正常(打印用法:./替换键并结束程序)
下面的代码
# include <stdio.h>
# include <cs50.h>
# include <string.h>
# include <stdlib.h>
# include <ctype.h>
int main(int argc, string argv[])
{
// Check that only one argument submitted
if (argc == 2)
{
// Check that key contains 26 characters
int keylen = strlen(argv[1]);
if (keylen == 26)
{
// Check that all characters are letters
for (int i = 0; i < keylen; i++)
{
bool lettercheck = isalpha(argv[1][i]);
if (lettercheck == true)
{
// THIS IS CAUSING ERROR - Check that no letters have been repeated - put all in lowercase to do so
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1][i - 1];
if (tolower(currentletter) == tolower(previousletter))
{
printf("Usage: ./substitution key
");
return 1;
}
}
}
else
{
printf("Usage: ./substitution key
");
return 1;
}
}
}
else
{
printf("Key must contain 26 characters.
");
return 1;
}
}
else
{
printf("Usage: ./substitution key
");
return 1;
}
// Get user input
string input = get_string("plaintext: ");
//Transform input using key
for(int i = 0; i < strlen(input); i++)
{
char currentletter = input[i];
int testlower = islower(currentletter);
int testupper = isupper(currentletter);
if (testupper > 0)
{
int j = input[i] - 65;
input[i] = toupper(argv[1][j]);
}
else if (testlower > 0)
{
int j = input[i] - 97;
input[i] = tolower(argv[1][j]);
}
}
printf("ciphertext: %s
", input);
}
编辑: 已找到解决方案-问题在于第二个for循环迭代了i-1次,而不是n次
代码应该是
charpreviouslletter = argv[1][n]
而不是
charpreviousletter = argv[1][i - 1]
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1]**[i - 1]**
在此循环中-
// THIS IS CAUSING ERROR - Check that no letters have been repeated - put all in lowercase to do so
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1][i - 1];
if (tolower(currentletter) == tolower(previousletter))
{
printf("Usage: ./substitution key
");
return 1;
}
}
您只将当前字符与前一个字符进行比较。这不适用于abcdefca
c
和a
有副本,但它们不在原始副本旁边,因此您的逻辑不会找到这些副本。您的逻辑仅适用于彼此相邻的重复项,如aabcddef
。
相反,您需要记下在遍历时遇到了哪些字符。如果您遇到已经遇到的字符,则知道有重复的字符。
值得庆幸的是,密钥只需要包含字母表的所有26个字符,没有任何重复。这意味着您可以简单地拥有26个槽的int
数组-每个槽计算字母在该索引处出现的次数。第0个索引代表‘a’,第1个索引代表‘b’,依此类推。
通过这种方式,您可以非常轻松地使用letter - 'a'
获取字母字符的索引,其中letter
是字母字符。因此,如果letter
为a
,则会得到0,这实际上是'a'
的索引此外,在遍历键时有一个嵌套循环,这个嵌套循环也会遍历键。除非它只在某个索引之前执行,该索引是外部循环的当前索引。这看起来既浪费又奇怪。为什么不简单地循环一次,检查当前字符是否是字母和还检查以前是否遇到过这个字母。这就是你要做的全部事情!
int letter_presence[26];
char upperletter;
string key = argv[1];
if (strlen(key) == KEY_LEN)
{
for (int index = 0; index < KEY_LEN; index++)
{
if (!isalpha(key[index]))
{
// Wrong key - invalid character
printf("Usage: ./substitution key
");
return 1;
}
if (letter_presence[tolower(key[index]) - 'a'] == 0)
{
// This letter has not been encountered before
letter_presence[upperletter - 'A'] = 1;
}
else
{
// Wrong key - Duplicate letters
return 1;
}
}
// All good
}
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