从网上找到的一些字符串函数源码,对比如下:
1. strlen(),计算字符串长度
int strlen (const char *s)
{
const char *p;
for (p = s; *p; ++p);
return p - s;
}
int strlen(const char *string)
{
int i=0;
while(string[i]) i++; //说明'\0'的ASIC码为0
return i;
}
2. strcpy(), 字符串拷贝.char *strcpy (char *dst, const char *src)
{
while ((*dst++ = *src++) != '/0');
return dst - 1;
}
char *strcpy(char *destination, const char *source)
{
while(*destinaton++=*source++); //指针指向空的时候为0
return (destination-1);
}
3. strcat(), 字符串的连接.
char *strcat (char *dst, const char *src)
{
char *ret = dst;
while (*dst++ != '/0');
--dst;
while ((*dst++ = *src++) != '/0');
return ret;
}
char *strcat(char *target,const char *source)
{
char *original=target;
while(*target) target++; // Find the end of the string
while(*target++=*source++);
return(original);
}
4. streql(), 判断两个字符串是否相等.int streql(char *str1,char *str2)
{
while((*str1==*str2)&&(*str1))
{
str1++;
str2++;
}
return((*str1==NULL)&&(*str2==NULL));
}
int streql(char *str1,char *str2)
{
while((*str1==*str2)&&(*str1))
{
str1++;
str2++;
}
return((*str1==NULL)&&(*str2==NULL));
}
5. strchr(), 在字符串中查找某个字符.char *strchr (const char *s, int c)
{
for (; *s != (char) c; ++s)
if (*s == '/0')
return NULL;
return (char *) s;
}
char *strchr(const char *string,int letter)
{
while((*string!=letter)&(*string))
string++;
return (string);
}
6. chrcnt(), 计算某个字符在字符串中出现的次数.int chrcnt(const char *string,int letter)
{
int count=0;
while(*string)
{
if(*string==letter)
count++;
string++;
}
return count;
}
int chrcnt(const char *string,int letter)
{
int count=0;
while(*string)
if(*string==letter)count++;
return count;
}
7. strcmp(), 判断两个字符串是否相等.int strcmp (const char *s1, const char *s2)
{
int ret;
while ((ret = *(unsigned char *) s1 - *(unsigned char *) s2++) == 0 && *s1++);
return ret;
}
int strcmp(const char *str1,const char *str2)
{
while((*str1==*str2)&&(*str1))
{
str1++;
str2++;
}
if((*str1==*str2)&&(!*str1)) //Same strings
return 0;
else if((*str1)&&(!*str2)) //Same but str1 longer
return -1;
else if((*str2)&&(!*str1)) //Same but str2 longer
else return((*str1>*str2)?-1:1);
}