Question

如何将字数和字符数返回到一个数组中？

#define NUMBER_MOVIES 3
void getWordCount( char movies[NUMBER_MOVIES][40]);


int main ()
{

    char movies[NUMBER_MOVIES][40] = {"Jurasic World", "Captain America", "I spite on your grave"};

    getWordCount(movies);


    return 0;
}

void getWordCount(char movies[NUMBER_MOVIES][40])
{
    int i;
    int wordCount = 0;

    int wordCharCount[ ]; // need to store to this array

    for(i = 0; i < strlen(movies[i]); i++)
    {
        if (*movies[i] == ' ')
        {
            wordCount++;
            printf ("word count :%s, %d \n", movies[i], wordCount);
        }

        printf ("word count :%s, %d \n", movies[i], wordCount);
    }

}

我想将所有单词计数和字符计数存储到同一数组中，即wordCharCount []

例如：侏罗纪世界；数组应为[2 8 5]。因此2个单词和第一个单词包含8个字符，第二个单词包含5个字符。另外，应该可以从函数外部访问array。

Answer 1

最明显的问题是您的字数只有一个循环，但是您需要扫描每部电影中的每个字符-这建议两个嵌套循环。

那么在单个数组中返回两种信息的想法是不明智的-结构会更合适。然后，在C语言中，您无法按值传递或返回数组，因此通常使调用者按引用传递数组以使函数放置结果。

考虑：

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

struct sMovie
{
    char title[40] ;
    int title_stats[40] ;
} ;

void getWordCount( struct sMovie* movies, int number_of_movies );

int main ()
{
    struct sMovie movies[] = { {"Jurassic World"}, 
                               {"Captain America"}, 
                               {"I spit on your grave"} } ;

    const int number_of_movies = sizeof(movies) / sizeof(*movies) ;
    getWordCount( movies, number_of_movies ) ;

    for( int m = 0; m < number_of_movies; m++ )
    {
        printf( "\"%s\" has %d words of lengths",
        movies[m].title,
        movies[m].title_stats[0] ) ;
        for( int w = 1; w <= movies[m].title_stats[0]; w++ )
        {
            printf( " %d", movies[m].title_stats[w] ) ;
        }
        printf( "\n" ) ;
    }

    return 0;
}

void getWordCount( struct sMovie* movies, int number_of_movies )
{
    // For each movie...
    for( int m = 0; m < number_of_movies; m++)
    {
        memset( movies[m].title_stats, 0, sizeof(movies[m].title_stats) ) ;
        movies[m].title_stats[0] = 1 ;

        // For each character
        for( int i = 0; movies[m].title[i] != 0; i++ )
        {
            int word = movies[m].title_stats[0] ; 
            movies[m].title_stats[word]++ ;

            if( movies[m].title[i] == ' ' )
            {
                movies[m].title_stats[word]-- ;
                movies[m].title_stats[0]++ ;
            }
        }
    }
}

输出：

"Jurassic World" has 2 words of lengths 8 5                                                                                                              
"Captain America" has 2 words of lengths 7 7                                                                                                             
"I spit on your grave" has 5 words of lengths 1 4 2 4 5

如何计算单词总数和给定字符串中每个单词的字符数并创建一维数组？

1 个答案: