计算二维字符串中的所有元音

Question

你好，我试着计算二维字符串中的所有元音。我的程序一直给我错误的输出，我不知道哪里出了问题。这是我的代码：

int main() {
    char strings[3][50] = { "hello WORLD", "hELLO", "Hello" };
    printf("%d \n", vowels_count_2D(3, 50, strings));

    return 0;
}

int vowels_count_2D(const int rows, const int cols, char string[][cols]) {
    for (int i = 0; i < rows; i++) {
        int vowels = string[i][0];
        for (int j = 0; j < cols; j++) {
            if (string[i][j] == 'a' || string[i][j] == 'e' ||
                string[i][j] == 'i' || string[i][j] == 'o' ||
                string[i][j] == 'u' || string[i][j] == 'A' ||
                string[i][j] == 'E' || string[i][j] == 'I' ||
                string[i][j] == 'O' || string[i][j] == 'U') {
                vowels++;
            }
        }
        return vowels;
    }
    return 0;
}

输出应为 7。我的输出是 107。为什么？

非常感谢。

Answer 1

循环逻辑不正确：vowels应该在外循环外初始化为0，return vowels;语句移到循环体外。

这是修改后的版本：

#include <stdio.h>

int vowels_count_2D(const int rows, const int cols, const char string[][cols]) {
    int vowels = 0;
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            if (string[i][j] == 'a' || string[i][j] == 'e' ||
                string[i][j] == 'i' || string[i][j] == 'o' ||
                string[i][j] == 'u' || string[i][j] == 'A' ||
                string[i][j] == 'E' || string[i][j] == 'I' ||
                string[i][j] == 'O' || string[i][j] == 'U') {
                vowels++;
            }
        }
    }
    return vowels;
}

int main() {
    char strings[3][50] = { "hello WORLD", "hELLO", "Hello" };
    printf("%d\n", vowels_count_2D(3, 50, strings));
    return 0;
}

与其扫描整个二维数组，不如在空终止符或每个字符串处停止：

#include <string.h>

int vowels_count_2D(const int rows, const int cols, char string[][cols]) {
    int vowels = 0;
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols && string[i][j] != '[=11=]'; j++) {
            if (strchr("aeiouAEIOU", string[i][j])) {
                vowels++;
            }
        }
    }
    return vowels;
}

Answer 2

My output is 107. Why?

您只计算了一行（由于过早 return vowels）但从 h 的 ASCII 值开始。您应该将 vowels 计数器设置为 0。

int vowels_count_2D(int rows, int cols,char string[][cols]) {
    int vowels = 0;
    for (int i=0; i<rows; i++) {

        printf("V: %d\n", vowels);

        for(int j=0; j<cols; j++) {
            if(string[i][j] =='a'|| string[i][j]=='e'||string[i][j]=='i'||string[i]
            [j]=='o'||string[i][j]=='u'||string[i][j]=='A'||string[i][j]=='E'||string[i]
            [j]=='I'||string[i][j]=='O' ||string[i][j]=='U' )
            {
                vowels++;
            }
        }
    }
    return vowels;
}

计算二维字符串中的所有元音

Count all vowels in 2D string

c

string

2d

count