itoa函数是怎么编写的？

[h_5_hao]

我在msdn上看见itoa这个函数 我想知道这个函数是怎么编写的 ~

[lbh2001]

在SDK的源代码里找找

[猫已经找不回了]

mark,itoa的源码不知道效率如何.

[xiaoyisnail]

转一个看看

/***
*atox.c - atoi and atol conversion
*
* Copyright (c) 1989-1997, Microsoft Corporation. All rights reserved.
*
*Purpose:
* Converts a character string into an int or long.
*
*******************************************************************************/

#include <cruntime.h>
#include <stdlib.h>
#include <ctype.h>

/***
*long atol(char *nptr) - Convert string to long
*
*Purpose:
* Converts ASCII string pointed to by nptr to binary.
* Overflow is not detected.
*
*Entry:
* nptr = ptr to string to convert
*
*Exit:
* return long int value of the string
*
*Exceptions:
* None - overflow is not detected.
*
*******************************************************************************/

long __cdecl atol(
const char *nptr
)
{
int c; /* current char */
long total; /* current total */
int sign; /* if ''-'', then negative, otherwise positive */

/* skip whitespace */
while ( isspace((int)(unsigned char)*nptr) )
++nptr;

c = (int)(unsigned char)*nptr++;
sign = c; /* save sign indication */
if (c == ''-'' || c == ''+'')
c = (int)(unsigned char)*nptr++; /* skip sign */

total = 0;

while (isdigit(c)) {
total = 10 * total + (c - ''0''); /* accumulate digit */
c = (int)(unsigned char)*nptr++; /* get next char */
}

if (sign == ''-'')
return -total;
else
return total; /* return result, negated if necessary */
}


/***
*int atoi(char *nptr) - Convert string to long
*
*Purpose:
* Converts ASCII string pointed to by nptr to binary.
* Overflow is not detected. Because of this, we can just use
* atol().
*
*Entry:
* nptr = ptr to string to convert
*
*Exit:
* return int value of the string
*
*Exceptions:
* None - overflow is not detected.
*
*******************************************************************************/

int __cdecl atoi(
const char *nptr
)
{
return (int)atol(nptr);
}

#ifndef _NO_INT64

__int64 __cdecl _atoi64(
const char *nptr
)
{
int c; /* current char */
__int64 total; /* current total */
int sign; /* if ''-'', then negative, otherwise positive */

/* skip whitespace */
while ( isspace((int)(unsigned char)*nptr) )
++nptr;

c = (int)(unsigned char)*nptr++;
sign = c; /* save sign indication */
if (c == ''-'' || c == ''+'')
c = (int)(unsigned char)*nptr++; /* skip sign */

total = 0;

while (isdigit(c)) {
total = 10 * total + (c - ''0''); /* accumulate digit */
c = (int)(unsigned char)*nptr++; /* get next char */
}

if (sign == ''-'')
return -total;
else
return total; /* return result, negated if necessary */
}

#endif /* _NO_INT64 */


#include <msvcrt/errno.h>
#include <msvcrt/stdlib.h>
#include <msvcrt/internal/file.h>
char* _itoa(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;
}

[lnever]

[Quote=引用 1 楼 sunjane2 的回复:]

void itoa(int n,char s[])//没考虑2、16进制
//考虑char s[]空间够用了吗？
{
int i,j,k,sign;
char tmp;
if((sign=n) <0)
      n=-n;
        i=0;
do{
        s[i++]=n%10+'0';
  }
while ((n/=10)>0);
        if(sign <0)
                s[i++]='-';
s[i]='\0';
//printf("debug int n: %d\n",n);
//printf("debug char s: %s\n",s);
/*
for debug input int i and output char s[];
*/
for(j=i-1,k=0;j>((i-1)/2);j--,k++)
{
      tmp=s[k];
      s[k]=s[j];
      s[j]=tmp;
}
//printf("debug str: %s\n",s);
} 

[/Quote]

[sunjane2]

我这个和标准的itoa()有点不一样，只是参数不一样，自己去看吧
void itoa(int n,char s[])
{
int i,j,k,sign;
char tmp;
if((sign=n)<0)
n=-n;
i=0;
do{
s[i++]=n%10+'0';
}
while ((n/=10)>0);
if(sign<0)
s[i++]='-';
s[i]='\0';
//printf("debug int n: %d\n",n);
//printf("debug char s: %s\n",s);
/*
for debug input int i and output char s[];
*/
for(j=i-1,k=0;j>((i-1)/2);j--,k++)
{
tmp=s[k];
s[k]=s[j];
s[j]=tmp;
}
//printf("debug str: %s\n",s);
}

[waizqfor]

转一个高效方法

 char   *myitoa(   int   value,   char   *str,   int   radix   )   
  {   
  static   char   szMap[]   =   {   
  '0',   '1',   '2',   '3',   '4',   '5',   
  '6',   '7',   '8',   '9',   'a',   'b',   
  'c',   'd',   'e',   'f',   'g',   'h',   
  'i',   'j',   'k',   'l',   'm',   'n',   
  'o',   'p',   'q',   'r',   's',   't',   
  'u',   'v',   'w',   'x',   'y',   'z'   
  }; //   字符映射表   
  int   nCount   =   -1,   nIndex;   
  char   *pStr   =   str,   nTemp;   
  if   (   radix   >=   2   &&   radix   <=   36   )   
  {   //   限制radix必须在2到36之间   
  if   (   value   <   0   &&   radix   ==   10   )   
  {   //   如果是负数就在首位添加负号，并将字符串前移   
  *pStr++   =   '-';   
  value   =   -value;   //转为正数，   
  }   
  unsigned   int   nValue   =   *(unsigned*)&value;   
  do   { //   循环转换每一个数字，直到结束   
  pStr[   ++nCount   ]   =   szMap[   nValue   %   radix   ];   
  nValue   /=   radix;   
  }   while(   nValue   >   0   );   //   转换结束后字符串是翻的   
  nIndex   =   (   nCount   +   1   )   /   2;   //   计算出一半的长度   
  while(   nIndex--   >   0   )   {   //   将字符串的字符序翻转   
  nTemp   =   pStr[   nIndex   ];   
  pStr[   nIndex   ]   =   pStr[   nCount   -   nIndex   ];   
  pStr[   nCount   -   nIndex   ]   =   nTemp;   
  }   
  }   
  pStr[   nCount   +   1   ]   =   '\0';   //   置结束符   
  return   str;   
  }