今天微薄上看到 http://weibo.com/1401880315/AaNkykg6J#_rnd1379945435634左耳朵耗子：“現在的程序員，連atoi()都不知道是什么了，沒事，那我改，不叫atoi()改叫StrToInt()，卻發現，好些人連ASCII碼都不知道是怎么一回事，沒事，我教會你。但最終卻發現怎么有這么多人連這樣簡單的程序都寫不好（包括有多年工作經驗的人）。“比技術更恐怖的是有一群不合格的程序員在使用這這些技術”。好久沒寫這么基本的代碼了，簡單嘗試寫了下， 結果花了半個小時，只寫了一個最基本功能的，通過后面的測試用例還花了不少時間進行調試修改。 INT StrToInt(const TCHAR* lpszValue)
{
assert(lpszValue != NULL);
INT nLen = lstrlen(lpszValue);
const TCHAR* pEnd = lpszValue + nLen;
TCHAR* pCurrent = (TCHAR*)lpszValue;
BOOL bNegative(FALSE);
if(nLen >= 1)
{
if(*pCurrent == _T('+'))
{
bNegative = FALSE;
pCurrent += 1;
}
else if(*pCurrent == _T('-'))
{
bNegative = TRUE;
pCurrent += 1;
}
else
{
bNegative = FALSE;
}
}
INT nBase(10);
if(pEnd - pCurrent >= 2)
{
if(pCurrent[0] == _T('0')
&& ::toupper(pCurrent[1]) == _T('X'))
{
pCurrent += 2;
nBase = 16;
}
}
INT nRet(0);
INT nValue(0);
while(pCurrent != pEnd)
{
TCHAR ch(*pCurrent);
if(ch >= _T('0') && ch <= _T('9'))
{
nValue = ch - _T('0');
}
else if(nBase == 16)
{
if(::toupper(ch) >= _T('A') && ::toupper(ch) <= _T('F'))
{
nValue = 10 + (::toupper(ch) - _T('A'));
}
else
{
assert(FALSE);
break;
}
}
else
{
assert(FALSE);
break;
}
nRet += nValue * pow((double)nBase, pEnd - pCurrent - 1);
pCurrent += 1;
}
return bNegative ? -nRet : nRet;
}
void test
{
assert(StrToInt(_T("11")) == 11);
assert(StrToInt(_T("+12")) == 12);
assert(StrToInt(_T("-123")) == -123);
assert(StrToInt(_T("-0x1CF")) == -0x1CF);
assert(StrToInt(_T("-0X123")) == -0x123);
assert(StrToInt(_T("0X123")) == 0x123);
}
感慨用慣了Windows API和STL， 對于最基本的字符串處理代碼反而寫不好了。細想一下這個基本的東西確實不好寫， 實際上我是上面只是考慮了10進制和16進制， 沒有考慮其他進制，也沒有考慮小數，非法的字符串或是溢出等情況， 而真正工業級的庫要考慮所有的情況， 另外還要考慮轉換效率等問題。實際上CRT源碼中有這個函數的實現：/***
*wcstol, wcstoul(nptr,endptr,ibase) - Convert ascii string to long un/signed
* int.
*
*Purpose:
* Convert an ascii string to a long 32-bit value. The base
* used for the caculations is supplied by the caller. The base
* must be in the range 0, 2-36. If a base of 0 is supplied, the
* ascii string must be examined to determine the base of the
* number:
* (a) First char = '0', second char = 'x' or 'X',
* use base 16.
* (b) First char = '0', use base 8
* (c) First char in range '1' - '9', use base 10.
*
* If the 'endptr' value is non-NULL, then wcstol/wcstoul places
* a pointer to the terminating character in this value.
* See ANSI standard for details
*
*Entry:
* nptr == NEAR/FAR pointer to the start of string.
* endptr == NEAR/FAR pointer to the end of the string.
* ibase == integer base to use for the calculations.
*
* string format: [whitespace] [sign] [0] [x] [digits/letters]
*
*Exit:
* Good return:
* result
*
* Overflow return:
* wcstol -- LONG_MAX or LONG_MIN
* wcstoul -- ULONG_MAX
* wcstol/wcstoul -- errno == ERANGE
*
* No digits or bad base return:
* 0
* endptr = nptr*
*
*Exceptions:
* Input parameters are validated. Refer to the validation section of the function.
*
*******************************************************************************/
/* flag values */
#define FL_UNSIGNED 1 /* wcstoul called */
#define FL_NEG 2 /* negative sign found */
#define FL_OVERFLOW 4 /* overflow occured */
#define FL_READDIGIT 8 /* we've read at least one correct digit */
static unsigned long __cdecl wcstoxl (
_locale_t plocinfo,
const wchar_t *nptr,
const wchar_t **endptr,
int ibase,
int flags
)
{
const wchar_t *p;
wchar_t c;
unsigned long number;
unsigned digval;
unsigned long maxval;
_LocaleUpdate _loc_update(plocinfo);
/* validation section */
if (endptr != NULL)
{
/* store beginning of string in endptr */
*endptr = nptr;
}
_VALIDATE_RETURN(nptr != NULL, EINVAL, 0L);
_VALIDATE_RETURN(ibase == 0 || (2 <= ibase && ibase <= 36), EINVAL, 0L);
p = nptr; /* p is our scanning pointer */
number = 0; /* start with zero */
c = *p++; /* read char */
while ( _iswspace_l(c, _loc_update.GetLocaleT()) )
c = *p++; /* skip whitespace */
if (c == '-') {
flags |= FL_NEG; /* remember minus sign */
c = *p++;
}
else if (c == '+')
c = *p++; /* skip sign */
if (ibase == 0) {
/* determine base free-lance, based on first two chars of
string */
if (_wchartodigit(c) != 0)
ibase = 10;
else if (*p == L'x' || *p == L'X')
ibase = 16;
else
ibase = 8;
}
if (ibase == 16) {
/* we might have 0x in front of number; remove if there */
if (_wchartodigit(c) == 0 && (*p == L'x' || *p == L'X')) {
++p;
c = *p++; /* advance past prefix */
}
}
/* if our number exceeds this, we will overflow on multiply */
maxval = ULONG_MAX / ibase;
for (;;) { /* exit in middle of loop */
/* convert c to value */
if ( (digval = _wchartodigit(c)) != -1 )
;
else if ( __ascii_iswalpha(c))
digval = __ascii_towupper(c) - L'A' + 10;
else
break;
if (digval >= (unsigned)ibase)
break; /* exit loop if bad digit found */
/* record the fact we have read one digit */
flags |= FL_READDIGIT;
/* we now need to compute number = number * base + digval,
but we need to know if overflow occured. This requires
a tricky pre-check. */
if (number < maxval || (number == maxval &&
(unsigned long)digval <= ULONG_MAX % ibase)) {
/* we won't overflow, go ahead and multiply */
number = number * ibase + digval;
}
else {
/* we would have overflowed -- set the overflow flag */
flags |= FL_OVERFLOW;
if (endptr == NULL) {
/* no need to keep on parsing if we
don't have to return the endptr. */
break;
}
}
c = *p++; /* read next digit */
}
--p; /* point to place that stopped scan */
if (!(flags & FL_READDIGIT)) {
/* no number there; return 0 and point to beginning of
string */
if (endptr)
/* store beginning of string in endptr later on */
p = nptr;
number = 0L; /* return 0 */
}
else if ( (flags & FL_OVERFLOW) ||
( !(flags & FL_UNSIGNED) &&
( ( (flags & FL_NEG) && (number > -LONG_MIN) ) ||
( !(flags & FL_NEG) && (number > LONG_MAX) ) ) ) )
{
/* overflow or signed overflow occurred */
errno = ERANGE;
if ( flags & FL_UNSIGNED )
number = ULONG_MAX;
else if ( flags & FL_NEG )
number = (unsigned long)(-LONG_MIN);
else
number = LONG_MAX;
}
if (endptr != NULL)
/* store pointer to char that stopped the scan */
*endptr = p;
if (flags & FL_NEG)
/* negate result if there was a neg sign */
number = (unsigned long)(-(long)number);
return number; /* done. */
}
透過這道題確實可以投射出一個程序員的計算機基本功， 大家可以自己嘗試下實現這個函數， 看看自己的計算機基本功。
posted on 2013-09-23 22:39
Richard Wei 閱讀(2754)
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