C++ CS50无元音返回空数组

我被困在n0 v0w3ls exercise from CS50号公路上了.我想不出如何使用非空白输出.我完全搞不懂字符串/字符之间的数据类型转换，并将新值赋给输出字符.我try 了这么多组合，我再也不明白其中的区别了.现在，我的程序可以在调试器中运行，但输出为空.任何非常感谢的帮助；)

#include <cs50.h>
#include <stdio.h>
#include <string.h>

char replace(string input);

int main(int argc, string argv[])
{
    if (argc != 2)
    {
        printf("Incorrect number of arguments\n");
        return 1;
    }
    else
    {
        printf("%c \n", replace(argv[1]));
    }

}

char replace(string input)
{
    int n = strlen(input);
    //n+1 as I read the last char should be a '\0' to be recognised as a string
    char s[n+1];

    for (int i = 0; i < n; i++)
    {
        //printf("%c\n", input[i]);
        switch (input[i])
        {
            case 97:
                s[i] = '6' ;
                break;
            case 101:
                s[i] = '3';
                break;
            case 105:
                s[i] = '1';
                break;
            case 111:
                s[i] = '0';
                break;
            default:
                s[i] = input[i];
                break;
        }
    }
    return s[n];
}

其目的是用数字代替元音.我试过了

• 将输入作为字符串
• 在该字符串的每个字符上循环
• 如果字符是元音&gt；在我的输出字符串中替换它，否则保持原样
• 是否返回我的输出字符数组/字符串？

// Write a function to replace vowels with numbers
// Get practice with strings
// Get practice with command line
// Get practice with switch

#include <cs50.h>
#include <stdio.h>
#include <string.h>

string replace(string input);

int main(int argc, string argv[])
{
    if (argc != 2)
    {
        printf("Incorrect number of arguments\n");
        return 1;
    }
    else
    {
        printf("%s \n", replace(argv[1]));
    }

}

string replace(string input)
{
    int n = strlen(input);
    string s[n+1];

    for (int i = 0; i < n; i++)
    {
        //printf("%c\n", input[i]);
        switch (input[i])
        {
            case 'a':
                s[i] = "6" ;
                break;
            case 'e':
                s[i] = "3";
                break;
            case 'i':
                s[i] = "1";
                break;
            case 'o':
                s[i] = "0";
                break;
            default:
                s[i] = &input[i];
                break;
        }
    }
    s[n+1] = "\0";
    return s[n];
}