C文本格式化字符数量关闭
因此，我有一个根据规则进行格式化文本的分配，并最终打印出字符串中字符的数量（包括\ n和Space，除了\ 0以外的任何其他内容字符串的结尾）。
基本上，有效的输入以一个点结束，但我认为最终点之后我还有更多的空间。 我尝试了几种方法，例如循环，从字符串末尾开始用\ 0代替空格。 删除
编辑要求的要求是：
- 将双点（..）转换为新线路
- 多个空格，
- 确保以前没有一个空间到逗号或点，
- 以确保逗号或点后有单个空间。
- 不更改撇号之间的原始内容。
- 并验证仅在正确的位置（新行\段落）中有大写字母。
我们必须在主函数（不幸的是）中执行所有代码，而我的代码通常会在数量中误认为字符的数量（可能是对最后一个点之后的额外空间，
这是我的输入示例代码在计数字符时失败
the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style .
int main() {
char ans;
printf("*************** Welcome to the text cleaner ***************\n\n");
do
{
int length, i, j = 0;
int word, sentence, para, space;
char tin[601], tout[601], * dex, * pos;
printf("\nPlease enter text:\n");
gets_s(tin, 600);
length = strlen(tin);
dex = strchr(tin, '.'); //converts double dots to new line
while (dex != NULL)
{
if (tin[dex - tin + 1] == '.') {
tin[dex - tin + 1] = '\n';
}
dex = strchr(dex + 1, '.');
}
length = strlen(tin);
dex = strchr(tin, ' '); //converting multiple spaces to single space
while (dex != NULL)
{
while (dex != NULL && tin[dex - tin + 1] == ' ')
{
for (i = dex - tin + 1; i < strlen(tin); i++)
{
tin[i - 1] = tin[i];
}
dex = strchr(dex, ' ');
j++;
}
dex = strchr(dex + 1, ' ');
}
tin[length - j] = '\0';
j = 0;
dex = strchr(tin, '\n');
while (dex != NULL && tin[dex-tin+1] == ' ') //delets spaces in the beggining of new row
{
for (i = dex - tin + 1;i < strlen(tin);i++) {
tin[i] = tin[i + 1];
}
dex = strchr(dex + 1, '\n');
}
dex = strchr(tin, ','); //deletes space before comma
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i++)
{
tin[i] = tin[i+1];
}
dex = strchr(dex+1, ',');
}
dex = strchr(tin, '.'); //deletes space before dots
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i++)
{
tin[i] = tin[i+1];
}
dex = strchr(dex + 1, '.');
}
dex = strchr(tin, ','); // adds space after comma
while (dex != NULL && tin[dex - tin + 1] != ' ')
{
if (tin[dex - tin + 1] != '\n')
{
tin[strlen(tin) + 1] = '\0';
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin + 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex + 1, ',');
}
}
dex = strchr(tin, '.'); // adds space after dot
while (dex != NULL && tin[dex - tin + 1] != ' ')
{
tin[strlen(tin) + 1] = '\0';
if (tin[dex - tin + 1] == '\n')
{
dex = strchr(dex + 1, '.');
}
else
{
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin + 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex + 1, '.');
}
}
strcpy_s(tout, sizeof(tout), tin);
_strlwr_s(tout,sizeof(tout)); //copies and lowercasing the input string
dex = strchr(tin, '"');
if (dex != NULL) {
pos = strchr(dex + 1, '"');
while (dex != NULL)
{
for (i = dex - tin; i < pos - tin; i++) {
tout[i] = tin[i];
}
dex = strchr(pos + 1, '"');
if (dex)
{
pos = strchr(dex + 1, '"');
}
} //making sure that the letters in the quotes have't been lowercased
}
_strupr_s(tin, sizeof(tin));
dex = strchr(tout, '.');
pos = strchr(tin, '.');
while (dex != NULL && pos != NULL)
{
tout[dex - tout + 2] = tin[pos - tin + 2];
dex = strchr(dex + 1, '.');
pos = strchr(pos + 1, '.');
}
//CAPSLOCK
dex = strchr(tout, '.'); //deletes space before dots
while (dex != NULL)
{
if (tout[dex - tout - 1] == ' ')
{
for (i = dex - tout - 1; i < strlen(tout); i++)
{
tout[i] = tout[i+1];
}
}
dex = strchr(dex + 1, '.');
}
if (tout[0] == ' ') {
for (i = 0 ;i < strlen(tout); i++) {
tout[i] = tout[i + 1];
}
}//handeling single space in the beggining of the string
if (tout[0] >= 'a' && tout[0] <= 'z') {
tout[0] -= 32;
} //First letter always capital
word = 0;
sentence = 0;
para = 1;
space = 0;
length = strlen(tout);
for (i = 0; tout[i] != '\0';i++)
{
if (tout[i] == ' ' && tout[i + 1] != ' ')
word++;
}
dex = strchr(tout, '.');
while (dex != NULL)
{
sentence++;
dex = strchr(dex + 1, '.');
}
dex = strchr(tout, '\n');
while (dex != NULL)
{
space++;
para++;
word++;
dex = strchr(dex + 1, '\n');
}
//dex = strchr(tout, '-');
//while (dex != NULL)
//{
// word++;
// dex = strchr(dex + 1, '-');
//}
printf_s("\nText after cleaning:\n------------------------------------------------------------------------------------------------\n");
printf_s("%s\n\n", tout);
printf_s("characters: %d | words: %d | sentences: %d | paragraphs: %d\n------------------------------------------------------------------------------------------------\n",length, word, sentence, para);
printf_s("\nIf you want to clean another string press (y): ");
scanf_s(" %c", &ans, 1);
if (ans == 'y')
{
gets_s(tin, 600);
}
} while (ans =='y');
So I have an assignment to format text according to rules and eventually print out the number of characters in the string (including \n and spaces, anything but the \0 at the end of the string).
Basically, a valid input is ended with a dot but I think I have a few more whitespaces after the final dots.
I have tried several approaches such as loop that replaces spaces with \0 starting from the end of the string.
nothing seems to work though...
EDIT The requirements are:
- to convert double-dots(..) to a new line
- delete multiple spaces leaving only one,
- making sure that there isn't a space prior to a comma or a dot
- making sure that there is single space after a comma or dot.
- not changing the original content between apostrophes.
- and validating that there are Capital letters only in the correct places (new line\paragraph).
we are required to do all of the code in the main function (unfortunately) and me code usually mistake the characters count by 1-2 extra in the count (probably do to extra spaces after the last dot
this is an example of input that my code fails at counting characters
the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style .
int main() {
char ans;
printf("*************** Welcome to the text cleaner ***************\n\n");
do
{
int length, i, j = 0;
int word, sentence, para, space;
char tin[601], tout[601], * dex, * pos;
printf("\nPlease enter text:\n");
gets_s(tin, 600);
length = strlen(tin);
dex = strchr(tin, '.'); //converts double dots to new line
while (dex != NULL)
{
if (tin[dex - tin + 1] == '.') {
tin[dex - tin + 1] = '\n';
}
dex = strchr(dex + 1, '.');
}
length = strlen(tin);
dex = strchr(tin, ' '); //converting multiple spaces to single space
while (dex != NULL)
{
while (dex != NULL && tin[dex - tin + 1] == ' ')
{
for (i = dex - tin + 1; i < strlen(tin); i++)
{
tin[i - 1] = tin[i];
}
dex = strchr(dex, ' ');
j++;
}
dex = strchr(dex + 1, ' ');
}
tin[length - j] = '\0';
j = 0;
dex = strchr(tin, '\n');
while (dex != NULL && tin[dex-tin+1] == ' ') //delets spaces in the beggining of new row
{
for (i = dex - tin + 1;i < strlen(tin);i++) {
tin[i] = tin[i + 1];
}
dex = strchr(dex + 1, '\n');
}
dex = strchr(tin, ','); //deletes space before comma
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i++)
{
tin[i] = tin[i+1];
}
dex = strchr(dex+1, ',');
}
dex = strchr(tin, '.'); //deletes space before dots
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i++)
{
tin[i] = tin[i+1];
}
dex = strchr(dex + 1, '.');
}
dex = strchr(tin, ','); // adds space after comma
while (dex != NULL && tin[dex - tin + 1] != ' ')
{
if (tin[dex - tin + 1] != '\n')
{
tin[strlen(tin) + 1] = '\0';
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin + 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex + 1, ',');
}
}
dex = strchr(tin, '.'); // adds space after dot
while (dex != NULL && tin[dex - tin + 1] != ' ')
{
tin[strlen(tin) + 1] = '\0';
if (tin[dex - tin + 1] == '\n')
{
dex = strchr(dex + 1, '.');
}
else
{
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin + 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex + 1, '.');
}
}
strcpy_s(tout, sizeof(tout), tin);
_strlwr_s(tout,sizeof(tout)); //copies and lowercasing the input string
dex = strchr(tin, '"');
if (dex != NULL) {
pos = strchr(dex + 1, '"');
while (dex != NULL)
{
for (i = dex - tin; i < pos - tin; i++) {
tout[i] = tin[i];
}
dex = strchr(pos + 1, '"');
if (dex)
{
pos = strchr(dex + 1, '"');
}
} //making sure that the letters in the quotes have't been lowercased
}
_strupr_s(tin, sizeof(tin));
dex = strchr(tout, '.');
pos = strchr(tin, '.');
while (dex != NULL && pos != NULL)
{
tout[dex - tout + 2] = tin[pos - tin + 2];
dex = strchr(dex + 1, '.');
pos = strchr(pos + 1, '.');
}
//CAPSLOCK
dex = strchr(tout, '.'); //deletes space before dots
while (dex != NULL)
{
if (tout[dex - tout - 1] == ' ')
{
for (i = dex - tout - 1; i < strlen(tout); i++)
{
tout[i] = tout[i+1];
}
}
dex = strchr(dex + 1, '.');
}
if (tout[0] == ' ') {
for (i = 0 ;i < strlen(tout); i++) {
tout[i] = tout[i + 1];
}
}//handeling single space in the beggining of the string
if (tout[0] >= 'a' && tout[0] <= 'z') {
tout[0] -= 32;
} //First letter always capital
word = 0;
sentence = 0;
para = 1;
space = 0;
length = strlen(tout);
for (i = 0; tout[i] != '\0';i++)
{
if (tout[i] == ' ' && tout[i + 1] != ' ')
word++;
}
dex = strchr(tout, '.');
while (dex != NULL)
{
sentence++;
dex = strchr(dex + 1, '.');
}
dex = strchr(tout, '\n');
while (dex != NULL)
{
space++;
para++;
word++;
dex = strchr(dex + 1, '\n');
}
//dex = strchr(tout, '-');
//while (dex != NULL)
//{
// word++;
// dex = strchr(dex + 1, '-');
//}
printf_s("\nText after cleaning:\n------------------------------------------------------------------------------------------------\n");
printf_s("%s\n\n", tout);
printf_s("characters: %d | words: %d | sentences: %d | paragraphs: %d\n------------------------------------------------------------------------------------------------\n",length, word, sentence, para);
printf_s("\nIf you want to clean another string press (y): ");
scanf_s(" %c", &ans, 1);
if (ans == 'y')
{
gets_s(tin, 600);
}
} while (ans =='y');
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评论（2）
正如我在首席评论中提到的那样，这可以在带有状态变量的单个循环中完成。
一些假设：
..（转换为newline）时，它就启动了一个新段落，不幸的是，我不得不完全重构代码。它是注释的。我意识到您必须仅 使用
main。额外的功能仅用于调试，因此它们“不计数”：这是程序输出：
更新：
不客气！
我很高兴您能够将我的代码纳入您的代码。这是所有可能学习的情况中最好的。
我不确定我是否添加了足够的注释来使我的算法清晰，所以我已经清理了一点，并且正在辩论是否要发布它。它在下面发布。
根据您刚才所说的话，我[仍然]不确定是否可以创建您自己的功能。正如我说的那样，如果不更改算法， debug 函数可能会很好。
无论如何，这是更新的代码：
带有
-d的输出：As I mentioned in the top comments, this can be done in a single loop with state variables.
A few assumptions:
..(which is converted to a newline), it starts a new paragraphUnfortunately, I had to completely refactor the code. It is annotated. I realize you have to only use
main. The extra functions are merely for debug, so they "don't count":Here is the program output:
UPDATE:
You're welcome!
I'm glad you were able to incorporate my code into yours. That's the best of all possible scenarios for learning.
I wasn't sure I had added enough annotation to make my algorithm clear, so I had cleaned it up a bit and was debating whether to post it or not. It is posted below.
Based on what you just said, I'm [still] not sure whether creating your own functions would have been allowed. As I said, debug functions would [probably] be okay if they don't change the algorithm.
Anyway, here is the updated code:
Output with
-d:这是一个可能的解决方案。
希望这将表明您不需要所有重复。
我只用给出的示例对其进行了测试，可能仍然有边缘案例可能会破裂。您可能需要分配
buffer而不是使用特定值，但是您可能需要检查是否有可能导致扩展的输入。关于正确位置的首都，除了一行以外，原始内容中没有其他任何范围，因此没有段落的概念。因此，我在句子开始时就去了上限。
nb： op没有指定正确的输出是什么，鉴于该帖子的标题为“错误的角色计数”，因此这是基于要求的要求和一些洞察力的最佳猜测正如所指出的不是产生正确的结果）。
我认为这里的重点不是要修复OP的错误，而是要说明越来越接近或实现解决方案的变更方式。
这会产生：
Here is a possible solution.
Hopefully this will show that you don't need all of the repetition.
I've only tested it with the example given, there might well still be edge cases where it might break. You might want to allocate the
bufferrather than using a specific value, but you might need to check to see if there are inputs that might lead to expansion.Regarding capitals in the right places, there is no scope in the original for anything other than a single line, so no concept of paragraphs. Therefore I've gone for caps at the start of sentences.
NB: OP didn't specify what the correct output was, given that the post is titled "wrong character count" so this is a best guess based on the requirements and some insight from the OP's code (which might as stated wasn't producing the right result).
I don't think that the point here is to fix the OP's bugs, but to illustrate alterantive ways of getting closer too, or achieving, a solution.
This produces: