Question

我一直在试图弄清楚如何计算一个信号的每个单词中的元音和字符。例如

在hello there句中

hello : 5 characters, 2 vowels

there : 5 characters, 2 vowels。我已经看到了完整句子做同样事情的代码。但不是一个字一个字。

以下是我一直致力于的编码

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lower case。我很难搞清楚这一点。

在运行代码时，我没有让元音计数。我能够分开这句话。但元音计数没有发生。

Answer 1

一种相当简单的方法：

#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;
}

Answer 2

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 3

这听起来很像家庭作业......

这里有一些伪 - 代码＆lt; - 下面不会按原样运行。只是为了显示逻辑。

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，实现v + c等细微的细节会在打印时给出你的总字长等。

Answer 4

你可能做的是，当你遇到一个＆＃34;时，你可以打印字符和元音的数量。＆＃34;（空格）然后重置计数器。这样，您就可以找到句子中每个单词的字符和元音。

Answer 5

如果您理解在整个句子中执行此操作的逻辑，那么您也可以通过简单地将句子分解为单个单词并对每个单词应用相同的逻辑来用单个单词来完成。你可以使用这样的事实：单词用空格（或多个，可能）分隔，将句子分解为单词。

计算句子每个单词中的元音和字符

5 个答案: