手写代码必备手册(C版) —— 读书笔记 1、字符串 API

</pre>现场编写 <span style="font-family:LMMono10-Regular; font-size:9pt">strcpy, strlen, strstr, atoi<span style="font-family:AdobeSongStd-Light; font-size:9pt">等库函数<br style="orphans:2; widows:2" /></span></span><p></p><pre name="code" class="cpp">typedef  unsigned int  size_t

1、1 strlen

实现
strlen，获取字符串长度，函数原型如下：

size_t strlen(const char *str);

<span style="font-family: LMMono9-Regular;">size_t</span> strlen ( const char * str )
{
const char *eos = str;

while( *eos++ ) ;

return( eos - str - 1 );
}

1.2 strcpy

实现
strcpy，字符串拷贝函数，函数原型如下：

char* strcpy(char *to, const char *from);

char* strcpy(char *to, const char *from)
{
assert(to != NULL && from != NULL);
char *p = to;
while ((*p++ = *from++) != '\0')
;
return to;
}

<span style="font-family: Arial, Helvetica, sans-serif; font-size: 12px;">
</span>

<span style="font-family: Arial, Helvetica, sans-serif; font-size: 12px;"><span style="font-family: LinLibertineOZ; font-size: 11pt;">1.3 strstr<br style="orphans: 2; text-align: -webkit-auto; widows: 2;" />
</span></span><pre name="code" class="cpp">实现 strstr，子串查找函数，函数原型如下：
char * strstr(const char *haystack, const char *needle);

<span style="font-family: Arial, Helvetica, sans-serif; font-size: 12px;"></span>

char *  strstr ( const char * str1, const char * str2 )
{
char *cp = (char *) str1;
char *s1, *s2;

if ( !*str2 )
return((char *)str1);

while (*cp)
{
s1 = cp;
s2 = (char *) str2;

while ( *s1 && *s2 && !(*s1-*s2) )
s1++, s2++;

if (!*s2)
return(cp);

cp++;
}

return(NULL);

}

</pre><p></p><p><span style="font-family: LinLibertineOZ; font-size: 11pt;"></span></p><span style="font-family:LinLibertineOZ; font-size:11pt"><strong>1.4 atoi</strong><span style="font-family:AdobeHeitiStd-Regular; font-size:9pt"><span style="font-family:AdobeSongStd-Light; font-size:9pt">实现 <span style="font-family:LMMono10-Regular; font-size:9pt">atoi<span style="font-family:AdobeSongStd-Light; font-size:9pt">，将一个字符串转化为整数，函数原型如下：<span style="font-family:LMMono9-Regular; font-size:8pt">int atoi(const char *str);</span></span></span></span></span></span><p><span style="font-family:LinLibertineOZ; font-size:11pt"><span style="font-family:LinLibertineOZ; font-size:11pt"><span style="font-family:AdobeHeitiStd-Regular; font-size:9pt"><span style="font-family:AdobeSongStd-Light; font-size:9pt"><span style="font-family:LMMono10-Regular; font-size:9pt"><span style="font-family:AdobeSongStd-Light; font-size:9pt"><span style="font-family:LMMono9-Regular; font-size:8pt"><span style="font-family:AdobeHeitiStd-Regular; font-size:9pt">分析</span></span></span></span><br style="orphans:2; widows:2" /></span></span></span><span style="font-family:AdobeSongStd-Light; font-size:9pt">注意，这题是故意给很少的信息，让你来考虑所有可能的输入。</span></span></p><p></p><div style="orphans:auto; widows:1"><span style="font-family:AdobeSongStd-Light"></span></div><span style="font-family:LinLibertineOZ; font-size:11pt"><span style="font-family:AdobeSongStd-Light; font-size:9pt"><span style="font-family:AdobeSongStd-Light; font-size:9pt">注意几个测试用例：<span style="font-family:LinLibertineO; font-size:9pt">1. <span style="font-family:AdobeSongStd-Light; font-size:9pt">不规则输入，但是有效， <span style="font-family:LinLibertineO; font-size:9pt">”-3924x8fc”<span style="font-family:AdobeSongStd-Light; font-size:9pt">，<span style="font-family:LinLibertineO; font-size:9pt">” + 413”,<span style="font-size:9pt">2. <span style="font-family:AdobeSongStd-Light; font-size:9pt">无效格式， <span style="font-family:LinLibertineO; font-size:9pt">” ++c”, ” ++1”<span style="font-size:9pt">3. <span style="font-family:AdobeSongStd-Light; font-size:9pt">溢出数据， <span style="font-family:LinLibertineO; font-size:9pt">”2147483648”</span></span></span><br style="orphans:2; widows:2" /></span></span></span></span></span></span></span></span></span></span></span><pre name="code" class="cpp"><pre name="code" class="cpp">int atoi(const char *str)
{
int num = 0;
int sign = 1;
const int len = strlen(str);
int i = 0;
while (str[i] == ' ' && i < len) i++;
if (str[i] == '+') i++;
if (str[i] == '-')
{
sign = -1;
i++;
}
for (; i < len; i++)
{
if (str[i] < '0' || str[i] > '9')
break;
if (num > INT_MAX / 10 ||
(num == INT_MAX / 10 && (str[i] - '0') > INT_MAX % 10))
{
return sign == -1 ? INT_MIN : INT_MAX;
}
num = num * 10 + str[i] - '0';
}
return num * sign;
}