C语言中实现itoa函数的实例

char *itoa( int value, char *string,int radix);

#include <iostream> 
#include <string> 
using namespace std; 
 
char* My_itoa(int value,char str[],int radix) 
{ 
  char temp[33]; 
  char *tp = temp; 
  int i; 
  unsigned v; 
  int sign; 
  char *sp; 
  if(radix > 36 || radix < 1) 
    return 0; 
  sign = (radix == 10 && value < 0); //十进制负数 
  if(sign) 
    v = -value; 
  else 
    v = (unsigned)value; 
  while(v || tp == temp)       //转化操作 
  { 
    i = v % radix; 
    v = v / radix; 
    if(i < 10) 
      *tp++ = i + '0'; 
    else 
      *tp++ = i + 'a' - 10; 
  } 
  if(str == 0) 
    str = (char*)malloc((tp - temp) + sign + 1); 
  sp = str; 
  if(sign)   //是负数的话把负号先加入数组 
    *sp++ = '-'; 
  while(tp > temp) 
    *sp++ = *--tp; 
  *sp = 0; 
 
  return str; 
} 
 
int main() 
{ 
  long int num; 
  int radix;   //输入的几进制表示 
  char str[256]; 
  cout<<"请输入整数和进制数："; 
  cin>>num>>radix; 
  My_itoa(num,str,radix); 
  cout<<"整数转换为字符串后："; 
  cout<<str<<endl; 
 
  return 0; 
}