C-计数字符串中的单词

Question

我一直在尝试做一个计算C语言字符串中单词数量的函数。但是，在某些casas中（如示例中的那个），它应该返回0而不是1 ...关于可能的任何想法错误？

#import <stdio.h>  

int contaPal(char s[]) {          
    int r;     
    int i;     
    r = 0;      

    for (i = 0; s[i] != '\0'; i++) {          
        if (s[i] == '\n')           
            r = r + 0;               

        if (s[i] != ' ' && s[i + 1] == ' ' && s[i + 1] != '\0')             
            r++;        

        if (s[i] != ' ' && s[i + 1] == '\0') {              
            r++;        
        }       
    }          
    return r; 
}  

int main () {   
    char s[15] = { ' ', '\n', '\0' };

    printf("Words: %d \n", (contaPal(s)));
    return 0; 
}

Answer 1

您不应将'\\n'与其他空格字符区别对待。

这是一个简单的版本：

#include <ctype.h>  
#include <stdio.h>  

int contaPal(const char *s) {          
    int count = 0, hassep = 1;

    while (*s) {
        if (isspace((unsigned char)*s) {
            hassep = 1;
        } else {
            count += hassep;
            hassep = 0;
        }
        s++;
    }
    return count;
}

int main(void) {   
    char s[] = " \n";

    printf("Words: %d\n", contaPal(s));
    return 0; 
}

Answer 2

我想这个词是除空格字符之外的任何字符序列。

您的函数返回1，因为遇到这种情况时，对于提供的字符串，遇到换行符时，变量r会增加

    if (s[i] != ' ' && s[i + 1] == '\0') {              
        r++;        
    }  

因此，函数实现是错误的。

可以按照演示程序中所示的以下方式进行定义

#include <stdio.h>
#include <ctype.h>

size_t contaPal( const char s[] ) 
{
    size_t n = 0;

    while ( *s )
    {
        while ( isspace( ( unsigned char )*s ) ) ++s;
        n += *s != '\0';
        while ( *s && !isspace( ( unsigned char )*s ) ) ++s;
    }

    return n;
}  

int main(void) 
{
    char s[] = { ' ', '\n', '\0' };

    printf( "Words: %zu\n", contaPal( s ) );

    return 0;
}

如您所料，它的输出是

Words: 0

Answer 3

使用现有字符测试功能的简单图示：

int main(void)
{
    int cnt = 0;
    int numWords = 0;
    BOOL trap = 0; //start count only after seeing a word
    char *sentence = "This is a sentence, too long.";       
    //char *sentence2 = "      ";//tested for empty string also


    while (*sentence != '\0') 
    {
        if ( isalnum (*sentence) ) //word is found, set trap and start count
        {
            sentence++;  //alpha numeric character, keep going
            trap = 1;
        }
        else if ( (( ispunct (*sentence) ) || ( isspace(*sentence) )) && trap)
        {  //count is started only after first non delimiter character is found
            numWords++;
            sentence++;
            while(( ispunct (*sentence) ) || ( isspace(*sentence) ))
            { //handle sequences of word delimiters
                sentence++;
            }
        }
        else //make sure pointer is increased either way
        {
            sentence++;
        }

    }
    return 0;
}

Answer 4

该行：

    if (s[i] != ' ' && s[i + 1] == ' ' && s[i + 1] != '\0')             
        r++;

当您查看'\\n'时，完全匹配大小写。

您应该使用if ... else if ...