计算文本中单词的频率数[重复]

Question

我写了一个计算文本中特定单词频率的函数。这个程序每次都返回零。我该如何改进它？

while (fgets(sentence, sizeof sentence, cfPtr))
{
for(j=0;j<total4;j++)
        {
            frequency[j] = comparision(sentence,&w);
            all_frequency+=frequency[j];
}}
.
.
.
int comparision(const char sentence[ ],char *w)
{  
    int length=0,count=0,l=0,i;
    length= strlen(sentence);
    l= strlen(w);
    while(sentence[i]!= '\n')
    if(strncmp(sentence,w,l))
        count++;
    i++;
    return count;
    }

Answer 1

我已经校对了您的代码，并对编码风格和变量名称发表了评论。那里 仍然是我在条件中留下的一个缺陷，这是由于没有迭代 句子。

这是您标记的代码：

while(fgets(sentence, sizeof sentence, cfPtr)) {
    for(j=0;j<total4;j++){
        frequency[j] = comparision(sentence,&w);
        all_frequency+=frequency[j];
    }

}

// int comparision(const char sentence[ ],char *w)  w is a poor variable name in this case.

int comparison(const char sentence[ ], char *word)  //word is a better name.
{

    //int length=0,count=0,l=0,i;   

    //Each variable should get its own line.
    //Also, i should be initialized and l is redundant.
    //Here are properly initialized variables:

    int length = 0;
    int count = 0;
    int i = 0;

    //length= strlen(sentence);   This is redundant, as you know that the line ends at '\n'

    length = strlen(word);  //l is replaced with length.

    //while(sentence[i]!= '\n') 

    //The incrementor and the if statement should be stored inside of a block 
    //(Formal name for curley braces).

    while(sentence[i] != '\n'){
        if(strncmp(sentence, word, length) == 0)  //strncmp returns 0 if equal, so you       
            count++;                              //should compare to 0 for equality
        i++;
    }
    return count;
}