itoa原型(c语言)

来源:互联网 发布:英国人口老龄化数据 编辑:程序博客网 时间:2024/06/03 18:23

#include <errno.h>
#include <stdlib.h>
//#include <file.h>
#include "stdio.h"

char *myitoa(int value, char *string, int radix)
{
 char tmp[33];
 char *tp = tmp;
 int i;
 unsigned v;
 int sign;
 char *sp;
 
 if (radix > 36 || radix <= 1)
 {
 // __set_errno(EDOM);
  return 0;
 }
 
 sign = (radix == 10 && value < 0);
 if (sign)
  v = -value;
 else
  v = (unsigned)value;
 while (v || tp == tmp)
 {
  i = v % radix;
  v = v / radix;
  if (i < 10)
   *tp++ = i+'0';
  else
   *tp++ = i + 'a' - 10;
 }
 
 if (string == 0)
  string = (char *)malloc((tp-tmp)+sign+1);
 sp = string;
 
 if (sign)
  *sp++ = '-';
 while (tp > tmp)
  *sp++ = *--tp;
 *sp = 0;
 return string;
}

 

void main(void)
{
 int num = 54321123;
 int iadix = 16;
 char result[20] = "";
 myitoa(num,result,iadix);
 printf("%s",result);
}

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