Question

我制作了一个代码，我的目标是将输入单词的空格放在句子中。我需要用空格补充小词

像：

Three witches watched three watches
tch

输出：

Three wi es wa ed three wa es

我制作了这段代码：

#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;


for (i = 0; i < B; i++)
{
    for(j=0;j<S;j++)
    {
        if(small[j]!=big[i])
        {
            j=0;
            break;
        }
        if(small[j]=='\0')
        {
            while (i-(j-1)!=i)
            {
                i = i - j;
                big[i] = '\n';
                i++;
            }
        }
    }
}
puts(big);
}

Answer 1

首先，在您的例子中，您使用换行符＆＃39; \ n＆＃39;而不是空间。
考虑这个简单的例子：

#include<stdio.h>
#define S 8
#define B 50
void main() {
    char small[S] = {"ol"};
    char big[B] = {"my older gradmom see my older sister"};
    int i = 0, j = 0;
    int cpt = 0;
    int smallSize = 0; 

    // loop to retrieve smallSize
    for (i = 0; i < S; i++)
    {
        if (small[i] != '\0')
            smallSize++;
    }

    // main loop
    for (i = 0; i < B; i++)
    {
        // stop if we hit the end of the string
        if (big[i] == '\0')
            break;
        // increment the cpt and small index while the content of big and small are equal
        if (big[i] == small[j])
        {
            cpt++;
            j++;
        }
        // we didn't found the full small word
        else
        {
            j = 0;
            cpt = 0;
        }

        // test if we found the full word, if yes replace char in big by space
        if (cpt == smallSize)
        {
            for (int k = 0; k < smallSize; k++)
            {
                big[i-k] = ' ';
            }
            j = 0;
            cpt = 0;
        }   
    }

    puts(big);
}

首先需要检索小数组的实际大小完成后，下一步是查看内部＆＃34;大＆＃34;如果里面有小字。如果我们找到它，那么用空格替换所有这些字符。

如果你想用一个空格替换整个小单词，那么你需要适应这个例子！

我希望这有帮助！

Answer 2

一种可能的方法是使用指向string的指针，一个用于读取，一个用于写入。这将允许用单个空格替换任意数量的字符（来自small的字符）。并且你真的不想嵌套循环，但只有一个来处理来自big的每个字符。

最后但并非最不重要的是，除了独立环境（内核或嵌入式开发）之外，永远不应该使用void main()。代码可能变成：

#include <stdio.h>
#define S 8
#define B 50

int main() {                     // void main is deprecated...
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0;    // pointer to written back big

for (i = 0; i < B; i++)
{
    if (big[i] == 0) break;       // do not process beyond end of string
    if(small[j]!=big[i])
    {
        for(int l=0; l<j; l++) big[k++] = small[l];  // copy an eventual partial small
        big[k++] = big[i];    // copy the incoming character
        j=0;                  // reset pointer to small
        continue;
    }
    else if(small[++j] == 0)  // reached end of small
    {
        big[k++] = ' ';       // replace chars from small with a single space
        j = 0;                // reset pointer to small
    }
}
big[k] = '\0';
puts(big);
return 0;
}

甚至更好（不需要固定大小的字符串）：

#include <stdio.h>

int main() {                     // void main is deprecated...
char small[] = {"ol"};
char big[] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0;    // pointer to written back big

for (i = 0; i < sizeof(big); i++)
{
    if(small[j]!=big[i])
...

Answer 3

在C字符串中以空字符终止＆＃39; \ 0＆＃39;。您的代码在开头（B和S）定义了一个不确定的随机数，并迭代了那么多字符而不是字符串实际包含的确切字符数。您可以通过在while循环中测试字符串的内容来使用字符串终止的事实。

i = 0;
while (str[i]) {
  ...
  i = i + 1;
}

如果您更喜欢for循环，则可以将其写为for循环。

for (i = 0; str[i]; i++) {
  ...
}

您的代码不会将剩余字符串的内容移动到左侧。如果用一个字符ol替换两个字符，则必须将剩余字符向左移动一个字符。否则你会在字符串中有一个洞。

#include <stdio.h>

int main() {
  char small[] = "ol";
  char big[] = "my older gradmom see my older sister";
  int s; // index, which loops through the small string
  int b; // index, which loops through the big string
  int m; // index, which loops through the characters to be modified

  // The following loops through the big string up to the terminating
  // null character in the big string.
  b = 0;
  while (big[b]) {
    // The following loops through the small string up to the
    // terminating null character, if the character in the small
    // string matches the corresponding character in the big string.
    s = 0;
    while (small[s] && big[b+s] == small[s]) {
      // In case of a match, continue with the next character in the
      // small string.
      s = s + 1;
    }
    // If we are at the end of the small string, we found in the
    // big string.
    if (small[s] == '\0') {
      // Now we have to modify the big string.  The modification
      // starts at the current position in the big string.
      m = b;
      // First we have to put the space at the current position in the
      // big string.
      big[m] = ' ';
      // And next the rest of the big string has to be moved left. The
      // rest of the big string starts, where the match has ended.
      while (big[b+s]) {
        m = m + 1;
        big[m] = big[b+s];
        s = s + 1;
      }
      // Finally the big string has to be terminated by a null
      // character.
      big[m+1] = '\0';
    }
    // Continue at next character in big string.
    b = b + 1;
  }
  puts(big);
  return 0;
}

c string：如果在句子中找到一个单词，则输入''

3 个答案: