计算句子每个单词中的元音和字符

Question

我一直在试图弄清楚如何计算句子每个单词中的元音和字符。例如

在你好有句子

< code>hello : 5个字符，2个元音

<代码>有:5个字符，2个元音 。我见过完整句子做同样事情的代码。但不是一个字一个字地。

下面是我一直在做的编码

int main() {
    char str[512] = "hello there", word[256];
    int i = 0, j = 0, v, h;
    str[strlen(str)] = '\0';

    /* checking whether the input string is NULL */
    if (str[0] == '\0') {
        printf("Input string is NULL\n");
        return 0;
    }

    /* printing words in the given string */
    while (str[i] != '\0') {
        /* ' '  is the separator to split words */
        if (str[i] == ' ') 
        {
            for (h = 0; word[h] != '\0'; ++h)
            {
                if (word[h] == 'a' || word[h] == 'e' || word[h] == 'i' || word[h] == 'o' || word[h] == 'u')++v;
            }
            printf("\nVowels: %d", v);
            word[j] = '\0';

            printf("%s\n", word);
            j = 0;
        } 
        else 
        {
            word[j++] = str[i];
        }
        i++;
    }

    word[j] = '\0';

    /* printing last word in the input string */
    printf("%s\n", word);
    return 0;
}

输入将全部为小写。我很难弄清楚这一点。

在运行代码时，我没有得到元音计数。我能把句子分开。但元音计数没有发生。

Answer 1

试试这段代码。它可能会帮助你

#include<stdio.h>
int main() {
    char str[512] = "hello there", word[256];
    int i = 0, j = 0, v=0,h; // you didn't initialize v to 0
    str[strlen(str)] = '\0';

    /* checking whether the input string is NULL */
    if (str[0] == '\0') {
            printf("Input string is NULL\n");
            return 0;
    }

    /* printing words in the given string */
    while (str[i] != '\0') {
            /* ' '  is the separator to split words */
            if (str[i] == ' ' ) {
    for (h = 0; word[h] != '\0'; h++) {
    if (word[h] == 'a' || word[h] == 'e' || word[h] == 'i' || word[h] == 'o' || word[h] == 'u')
            v++;
    }
    printf("%s :", word);
    printf(" %d chracters,",strlen(word));
    printf(" %d Vowels.\n", v);

    j = 0; v=0;
    word[j] = '\0';
    } else {
            word[j++] = str[i]; 
            word[j] = '\0';
    }
    i++;
    }

    /* calculating vowels in the last word*/ // when NULL occurs, Wont enter into while loop.
    for (h = 0; word[h] != '\0'; h++) {
    if (word[h] == 'a' || word[h] == 'e' || word[h] == 'i' || word[h] == 'o' || word[h] == 'u')
    v++;
    }
    printf("%s :", word);
    printf(" %d chracters,",strlen(word));
    printf(" %d Vowels.\n", v);

    return 0;
}

Answer 2

这听起来很像家庭作业…

这是一些伪代码

int c = 0;
int v = 0;
for (int i = 0; i < lengthOfSentence; i++){
    if (stringName[i] == '\0') { //optionally '\n' may be more suitable
        return;
    }
    if (stringName[i] == ' '){
        print previousWord // + c, v in whatever format you want
        c = 0;
        v = 0;
    }
    if (stringName[i] == vowel) { //you can do this part like in your code
        word[v+c] = stringName[i]; //get current char and add to next slot
        v++;
    }
    else {
        word[v+c] = stringName[i];
        c++++;
    }

除此之外，它还有一些微小的细节，例如实现 V C 会在打印时为您提供总字长等。

Answer 3

一个相当简单的方法：

#include <stdio.h>

const char* s(int n)
{
    return n == 1? "" : "s";
}

void count (const char* str)
{
    for (int i = 0;;)
        for (int v = 0, w = i;;)
        {
            int len;
            char c = str[i++];
            switch (c)
            {
            case 'a': case 'e': case 'i': case 'o': case 'u':
                v++;
            default:
                continue;
            case ' ': case '\t': case '\n': case '\0':
                len = i - 1 - w;
                printf("'%.*s': %d character%s, %d vowel%s\n", len, str+w, len, s(len), v, s(v));
                if (c)
                    break;
                else
                    return;
            }
            break;
        }
}


int main(void)
{
    count("My words with vowels");
    return 0;
}