c – 简化我的程序,将数字从一个基地转换到另一个

我正在读初级C课程我收到了一个作业,告诉我写一个程序,它将二进制和十六进制之间的任意数字的任意数字转换为二进制和十六进制之间的另一个基数.我被要求使用单独的函数来转换和从基数10转换.这是为了帮助我们习惯使用数组. (我们已经覆盖了以前在课堂上的参考.)我已经把它改成了,但我很确定这不是我的意图：

#include <iostream>
#include <conio.h>
#include <cstring>
#include <cmath>

using std::cout;
using std::cin;
using std::endl;

int to_dec(char value[],int starting_base);
char* from_dec(int value,int ending_base);

int main() {
    char value[30];
    int starting_base;
    int ending_base;
    cout << "This program converts from one base to another,so long as the bases are" << endl
        << "between 2 and 16." << endl
        << endl;
input_numbers:
    cout << "Enter the number,then starting base,then ending base:" << endl;
    cin >> value >> starting_base >> ending_base;
    if (starting_base < 2 || starting_base > 16 || ending_base < 2 || ending_base > 16) {
        cout << "Invalid base(s). ";
        goto input_numbers;
    }
    for (int i=0; value[i]; i++) value[i] = toupper(value[i]);
    cout << "Base " << ending_base << ": " << from_dec(to_dec(value,starting_base),ending_base) << endl
        << "Press any key to exit.";
    getch();
    return 0;
}

int to_dec(char value[],int starting_base) {
    char hex[16] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    long int return_value = 0;
    unsigned short int digit = 0;
    for (short int pos = strlen(value)-1; pos > -1; pos--) {
        for (int i=0; i<starting_base; i++) {
            if (hex[i] == value[pos]) {
                return_value+=i*pow((float)starting_base,digit++);
                break;
            }
        }
    }
    return return_value;
}

char* from_dec(int value,int ending_base) {
    char hex[16] = {'0','F'};
    char *return_value = (char *)malloc(30);
    unsigned short int digit = (int)ceil(log10((double)(value+1))/log10((double)ending_base));
    return_value[digit] = 0;
    for (; value != 0; value/=ending_base) return_value[--digit] = hex[value%ending_base];
    return return_value;
}

我很确定这比它想要的更先进.你觉得我应该怎么做？

我基本上正在寻找两种答案：

一个简单的解决方案就像我老师可能预料到的一样的例子.
>关于如何改进代码的建议.

for (int i=0; i<starting_base; i++) {

char * value = 3012,int base = 4,

Number = 3 * 4^3 + 0 * 4^2 + 1 * 4^1 + 2 * 4^0 = 192 + 0 + 4 + 2 = 198

x = 0
  x = 4x + 3 = 3
  x = 4x + 0 = 12
  x = 4x + 1 = 49
  x = 4x + 2 = 198

  return x;

unsigned
to_dec( char * inputString,unsigned base )
{
  unsigned rv = 0; // return value
  unsigned c;      // character converted to integer

  for( char * p = inputString; *p; ++p ) // p iterates through the string
  {
    c = *p - hex[0];
    rv = base * rv + c;
  }

  return rv;
}