itoa()和atoi()/atol()的源码

/***
*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 
#include 
#include 

/***
*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 
#include 
#include 
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;
}

PS1: atoi()可以用二维数组的索引来实现，避免*10的循环。

PS2: itoa()采用返回堆指针的方式得到结果，不会造成memory leak和指针问题。但是不可以通过传入的char*string返回结果，会造成指针问题。（可以改写为char**string，因此微软的sdk中实现的方式有问题）

转自；http://shiwawa.blog.hexun.com/17044824_d.html