你现在有一堆错误代码。作为一个问题，它缺乏重点。我建议你写一点代码。测试一下重复一遍一旦你厌倦了手工测试，就写一些单元测试。
我将与您分享一个工作实现，以便您可以了解如何构建解决方案。如果需要，可以将其用于单元测试。使用几个switch语句来确定word1和word2是否满足您在其中一个注解中提到的要求。我故意把这个留给你做练习。
不要使用任意大小的50，而是使用最大的输出来调整字符串的大小，并使用snprintf()来避免错误时的缓冲区溢出。如你所知，上限大小只是使用一个常规的堆栈分配字符串，而不是处理malloc()和free()每个循环迭代。
按照惯例，符号常量都是大写的。使用蛇的情况下，而不是 Camel 和蛇的情况下混合。

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

#define MIN_LARGE_NUMBER 20
#define MAX_WORD_SIZE sizeof "negative seventy eight"
// if s is the empty string return "" otherwise " "
#define sep(s) " " + !*(s)

const char *small_numbers[] = {
    "zero",
    "one",
    "two",
    "three",
    "four",
    "five",
    "six",
    "seven",
    "eight",
    "nine",
    "ten",
    "eleven",
    "twelve",
    "thirteen",
    "fourteen",
    "fifteen",
    "sixteen",
    "seventeen",
    "eighteen",
    "nineteen",
};

const char *large_numbers[] = {
    "twenty", // MIN_LARGE_NUMBER
    "thirty",
    "forty",
    "fifty",
    "sixty",
    "seventy",
    "eighty",
    "ninety"
};

char *get_number_word(int number, char word[MAX_WORD_SIZE]) {
    const char *negative;
    if(number < 0) {
        negative = "negative";
        number *= -1;
    } else {
        negative  = "";
    }

    // number - MIN_LARGE_NUMBER deals with the offset in the table,
    // and / 10 converts it to an array index.  The alternative would
    // be to leave the two two entries empty with [2] = "tweenty",
    // "thirty", ...
    const char *word1 = number >= MIN_LARGE_NUMBER ?
        large_numbers[(number - MIN_LARGE_NUMBER) / 10] :
        "";
    // for large numbers we only want the 2nd word if the last digit
    // is non-zero.
    const char *word2 = number < MIN_LARGE_NUMBER ?
        small_numbers[number] :
        number % 10 ?
            small_numbers[number % 10] :
            "";

    snprintf(word, MAX_WORD_SIZE, "%s%s" "%s%s" "%s",
        negative,
        sep(negative),
        word1,
        sep(word1),
        word2
        );
    return word;
}

int flush() {
    for(;;) {
        int ch = getchar();
        if(ch == EOF || ch == '\n')
            return ch;
    }
}

int main(void) {
    char word[MAX_WORD_SIZE];
    for(;;) {
        printf("Enter a two-digit number: ");
        int num;
        int rv = scanf("%d", &num);
        if(rv == EOF || flush() == EOF)
            break;
        if(rv != 1 || num < -99 || num > 99) {
            fprintf(stderr, "invalid input\n");
            continue; // or break?
        }
        printf("%s\n", get_number_word(num, word));
    }
    printf("\n");
}

示例会话（注意输入2a;如果你想更严格，那么你需要检查下一个输入是\n，例如“%d%c”;但是“%d”无论如何都会跳过前导空格，所以如果你想要严格的输入验证，请使用fgets()读取一行，使用strtol()读取一个数字）：

Enter a two-digit number: hello world
invalid input
Enter a two-digit number: 2a
two
Enter a two-digit number: -1000000
invalid input
Enter a two-digit number: -100
invalid input
Enter a two-digit number: -99
negative ninety nine 
Enter a two-digit number: -1
negative one 
Enter a two-digit number: 0
zero 
Enter a two-digit number: 1
one
Enter a two-digit number: 19
nineteen
Enter a two-digit number: 20
twenty 
Enter a two-digit number: 21
twenty one 
Enter a two-digit number: 99
ninety nine 
Enter a two-digit number: 100
invalid input
Enter a two-digit number: ^D