Question

我正在尝试将用户输入的句子拆分为单词数组，以便稍后可以将这些单词分别作为字符串进行操作。代码正在编译，但在用户输入后只打印垃圾。我试过调试，但没有看到问题。有人可以帮我解决吗？

#include<stdio.h> 
#include <string.h>

 int main() {
 char str[1000];
 int i = 0;
 char rev[1000][1000];
 int r = 0;


puts("Enter text:");

gets(str);

int k, length = 0;
printf_s("So the words are:\n");
while (str[i] != '\0')
{

    if (str[i] == ' ') {
        k = i - length;

        do {
            rev[r][k] = (str[k]);
            k++;

        } while (str[k] != ' ');
        printf(" ");
        length = (-1);
        r++;
    }
    else if (str[i + 1] == '\0') {
        k = i - length;

        do {
            rev[r][k] = (str[k]);
            k++;
        } while (str[k] != '\0');
        length = 0;
        r++;
    }

    length++;
    i++;
}for (int r = 0; r < 1000; r++) 
        printf("%s ", rev[r]);
return 0;

}

Answer 1

如果您希望将字符串拆分为字符串数组，则应考虑strtok中的#include <string.h>函数。 strtok函数将拆分给定分隔符上的字符串。对于您的情况，它将是" "。

使用教程要点中的strtok示例：

#include <string.h>
#include <stdio.h>

int main(){
   char str[80] = "This is - www.tutorialspoint.com - website";//The string you wish to split
   const char s[] = "-";//The thing you want it to split from. But there is no need to this.
   char *token;//Storing the string

   /* get the first token */
   token = strtok(str, s);//Split str one time using the delimiter s

   /* walk through other tokens */
   while( token != NULL ) 
   {
      printf( " %s\n", token );//Print the string

      token = strtok(NULL, s);//Split the string again using the delimiter
   }

   return(0);
}

Answer 2

替换你的声明

char rev[1000][1000];

与

char * rev[1000];                  // We will need pointers only
int    i = 0;                      // Index to previous array

以及

之后的所有代码

    puts( "Enter text:" );

用这个：

    fgets( str, 998, stdin );      // Safe way; don't use gets(str)

    const char delim[] = ",; ";    // Possible delimiters - comma, semicolon, space
    char *word;

    /* Get the first word */
    word     = strtok( str, delim );
    rev[i++] = word; 

    /* Get the next words */
    while( word != NULL ) 
    {
       word     = strtok( NULL, delim );
       rev[i++] = word;
    }

    /* Testing */
    for (int r = 0; r < i - 1; r++) 
        printf( "%s\n", rev[r] );

    return 0
}

正如你所看到的，所有肮脏的工作都是通过strtok()函数（“string to tokens”）来完成的，这些函数遍历其他单词（“tokens”） ，将其识别为字符串delim中的一个或多个字符分隔。

Answer 3

像这样修复

#include <stdio.h> 

int main(void) {
    char str[1000];
    char rev[1000][1000];

    puts("Enter text:");
    fgets(str, sizeof str, stdin);//Use fgets instead of gets. It has already been abolished.

    int r = 0;
    int k = 0;

    for(int i = 0; str[i] != '\0'; ++i){
        if (str[i] == ' ' || str[i] == '\n'){//is delimiter
            if(k != 0){
                rev[r++][k] = '\0';//add null-terminator and increment rows
                k = 0;//reset store position
            }
        } else {
            rev[r][k++] = str[i];
        }
    }
    if(k != 0)//Lastly there was no delimiter
        rev[r++][k] = '\0';

    puts("So the words are:");
    for (int i = 0; i < r; i++){
        printf("%s", rev[i]);
        if(i < r - 2)
            printf(", ");
        else if(i == r - 2)
            printf(" and ");
    }

    return 0;
}

Answer 4

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int count_spaces(char *str)
{
  if (str == NULL || strlen(str) <= 0)
    return (0);

  int i = 0, count = 0;
  while (str[i])
    {
      if (str[i] == ' ')
    count++;
      i++;
    }

  return (count);
}

int count_char_from_pos(char *str, int pos)
{
  if (str == NULL || strlen(str) <= 0)
    return 0;

  int i = pos, count = 0;
  while (str[i] && str[i] != ' ')
    {
      count++;
      i++;
    }

  return count;
}

char **get_words(char *str)
{
  if (str == NULL || strlen(str) <= 0)
    {
      printf("Bad string inputed");
      return NULL;
    }

  int i = 0, j = 0, k = 0;
  char **dest;

    if ((dest = malloc(sizeof(char*) * (count_spaces(str) + 1))) == NULL
    || (dest[0] = malloc(sizeof(char) * (count_char_from_pos(str, 0) + 1))) == NULL)
      {
    printf("Malloc failed\n");
    return NULL;
      }

  while (str[i])
    {
      if (str[i] == ' ') {
    dest[j++][k] = '\0';
      if ((dest[j] = malloc(sizeof(char) * (count_char_from_pos(str, i) + 1))) == NULL)
        {
          printf("Malloc failed\n");
          return NULL;
        }
      k = 0;
    }
      else { 
      dest[j][k++] = str[i];
    }

      i++;
    }

  dest[j][k] = 0;
  dest[j + 1] = NULL;
  return dest;
}

int main(void) {
    char *line = NULL;
    size_t n = 0;
    getline(&line, &n, stdin);
    printf("%s\n", line);
    line[strlen(line) - 1] = 0;
    printf("%s\n", line);

    char **tab = get_words(line);

    int i = 0;
    while (tab[i])
      {
    printf("%s\n", tab[i++]);
      }
}

这是一个很长但完全有效的例子获取用户输入然后将其发送到get_words函数。它将获得单词的数量，每个单词的字符数，在内存中分配所有内容并写入字符然后返回它。你得到一个char **并打印它只是测试它的工作

将字符串拆分为字符串数组

4 个答案: