实现一些字符串操作标准库函数、解决一些字符串问题

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mob604756f4c9c3 2013-09-16 19:01:00

文章标签 字符串 i++ c++ #include 字符串反转 文章分类 MySQL 数据库

一、实现字符串操作标准库函数

（1）、strcpy、strncpy、memmove、memcpy、strlen、strncat 的实现

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// The  strcpy()  function  copies  the  string  pointed to by src, including the terminating null byte ('\0'), to the buffer
// pointed to by dest.  The strings may not overlap, and the destination string dest must be  large  enough  to  receive  thecopy.
char *strcpy( char *dest, const char *src)
{
    assert((src != NULL) && (dest != NULL));
    size_t i;
     for (i = 0; src[i] != '\0'; i++)
        dest[i] = src[i];
    dest[i] = '\0';
     return dest;
}

// The strncpy() function is similar, except that at most n bytes of src are copied.  Warning: If there is no null byte among the first n bytes of src,
// the string placed in dest will not be null-terminated. If the length of src is less than n, strncpy() pads the remainder of dest with null bytes.
char *strncpy( char *dest, const char *src, size_t n)
{
    assert((src != NULL) && (dest != NULL));
    size_t i;
     for (i = 0; i < n && src[i] != '\0'; i++)
        dest[i] = src[i];
     for (; i < n; i++)
        dest[i] = '\0';
     return dest;
}

/* 借助于一个临时缓冲区temp ,即使src 和dest 所指的内存区间有重叠也能正确拷贝。*/
void *memmove( void *dest, const void *src, size_t n)
{
    assert((src != NULL) && (dest != NULL));
     char temp[n];
     int i;
     char *d = dest;
     const char *s = src;

     for (i = 0; i < n; i++)
        temp[i] = s[i];
     for (i = 0; i < n; i++)
        d[i] = temp[i];

     return dest;
}

/* 在32位的x86平台上,每次拷贝1个字节需要一条指令,每次拷贝4个字节也只需要一条指
* 令,memcpy函数的实现尽可能4个字节4个字节地拷贝 */
void *memcpy( void *dest, const void *src, size_t n)
{
    assert((src != NULL) && (dest != NULL));
     char *d = dest;
     const char *s = src;
     int *di;
     const int *si;
     int r = n % 4;
     while (r--)
        *d++ = *s++;
    di = ( int *)d;
    si = ( const int *)s;
    n /= 4;
     while (n--)
        *di++ = *si++;
     return dest;
}

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size_t strlen( const char *p)
{

assert(p != NULL);
    size_t size = 0;
     while (*p++ != '\0')
        ++size;
     return size;
}

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char *strncat( char *dest, const char *src, size_t n)
{
    size_t dest_len = strlen(dest);
    size_t i;

     for (i = 0 ; i < n && src[i] != '\0' ; i++)
        dest[dest_len + i] = src[i];
    dest[dest_len + i] = '\0';

     return dest;
}

二、解决字符串问题

（1）、将单词之间出现一个或多个连续的空白字符都压缩为1个。

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//编一个函数，输入一个字符串，要求做一个新字符串，把其中所有的一个或多个连续的空白字符都压缩为一个空格。这里所说的空白包括空格、'\t'、'\n'、'\r'。
char *shrink_space( char *dest, const char *src, size_t n)
{
    assert((src != NULL) && (dest != NULL));
    size_t i, j;

    dest[ 0] = src[ 0];

     for (i = 1, j = 1; src[i] != '\0'; i++, j++)
    {
         if (src[i] == '\t' || src[i] == '\n'
                || src[i] == '\r' || src[i] == ' ')

             if (src[i - 1] != '\t' && src[i - 1] != '\n'
                    && src[i - 1] != '\r' && src[i - 1] != ' ')

                dest[j] = ' ';
             else
                j--;
         else
            dest[j] = src[i];
    }
    dest[j] = '\0';

     return dest;
}

（2）、解析 URL 中的路径和查询字符串。? 号后面是查询字符串，由 “key=value” 形式的键值对组成，以 & 隔开。

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/*************************************************************************
    > File Name: find_url_token.c
    > Author: Simba
    > Mail: dameng34@163.com
    > Created Time: Sat 26 Jan 2013 04:05:32 PM CST
************************************************************************/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 10
typedef struct
{
     char *tokens[N];
     int count;
} unit_t;

void find_url_token( char str[], const char tok[], unit_t *ptr)
{
     int i;
     char *token = NULL;
     char *saveptr = NULL;

    ptr->count = 0;
     const char *needle = "://";

     if (strstr(str, needle))
    {
         for (i = 0; ; str = NULL, i++)
        {
            token = strtok_r(str, tok, &saveptr);
             if (token == NULL)
                 break;
             else
            {
                ptr->tokens[i] = token;
                ptr->count++;
            }
        }
    }
}

int main( void)
{
     /* 不能定义为char *url = "..."; 因为此时是定义一个指向字符串字面值(位于.rodata段)的指针,而
       调用strtok_r函数会修改这个字符串,运行时会产生段错误 */
     char url[] = "http://www.google.cn/search?complete=1&hl=zh-CN&ie=GB2312&q=linux&meta=";
     /* 给url初始化用的这个字符串并没有分配在.rodata段,而是直接写在指令里了,
     * 运行程序时通过三条movl 指令把12个字节写到栈上,这就是url的存储空间*/
    unit_t *ptr = malloc( sizeof(unit_t));
    find_url_token(url, "?&", ptr);
     int i;
     for (i = 0; i < ptr->count; i++)
        printf( "%s\n", ptr->tokens[i]);
    free(ptr);
     return 0;
}

（3）、去除\r\n，去除左右空白字符

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void str_trim_crlf( char *str)
{
     char *p = &str[strlen(str) - 1];
     while (*p == '\r' || *p == '\n')
        *p-- = '\0';

}

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void AllTrim( char *str )
{
     char *head, *tail;

     if ( str == NULL )
         return;

     for( head = str; *head == ' ' || *head == '\t'; head ++ );

     for( tail = str + strlen(str) - 1; (*tail == ' ' || *tail == '\t' ) && tail >= head; tail -- );

     while( head <= tail )
        *str ++ = *head ++;

    *str = 0;
}

（4）、判断字符串是否为回文

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// 判断字符串是否为回文
bool isSysmmetry( const char *src)
{
    assert(src != NULL);
     int len = strlen(src);
    assert(len != 0);
     const char *tmp = src + len - 1;

     int i;
     for (i = 0; i < len / 2; i++)
    {
         if (*src++ != *tmp--)
             break;
    }

     if (i == len / 2)
         return true;
     else
         return false;

}

（5）、google笔试：编码实现求给定字符串（全为小写英文字母）的最小后继，如 “abc” 的最小后继为“abd”, "dhz" 的最小后继为“di"。

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int MinNextStr( const char *src, char *&minnext)
{
     int srclen = strlen(src);
    minnext = ( char *)malloc((srclen + 1) * sizeof( char));
     if (minnext == NULL)
         return - 1;
    strcpy(minnext, src);

     int i = srclen - 1;
     while (i >= 0)
    {
        minnext[i]++;
         if (minnext[i] <= 'z')
             break;
        i--;
    }

     if (i < 0)
         return 0;
     else
    {
        minnext[++i] = '\0';
         return 1;

    }
}

如果把给定字符串全为小写英文字母改为大小写英文字母，则只要把第13行改为：

if (minnext[i] <= 'z' && minnext[i] >= 'a' || minnext[i] <= 'Z');

（6）、中兴：编码实现字符串右移n位，如“diopheg” 右移2位为“egdioph”。

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bool RightMoveStr( char *src, int n)
{
     int len = strlen(src);
     int mov = n % len;
     char *rstr = ( char *)malloc((mov + 1) * sizeof( char));
     if (rstr == NULL)
         return false;

     int i = 0;
     while (i < mov)
    {
        rstr[i] = src[len - mov + i];
        i++;
    }
    rstr[i] = '\0';

    i = len - mov - 1;
     while (i >= 0)
    {
        src[i + mov] = src[i];
        i--;
    }

    i = 0;
     while (i < mov)
    {
        src[i] = rstr[i];
        i++;
    }

    free(rstr);
     return true;
}

bool RightMove( char *src, char *&ssrc, int n)
{
     int len = strlen(src);
    ssrc = ( char *)malloc( sizeof( char) * (len + 1));
     if (ssrc == NULL)
         return false;

    n = n % len;
     char *tmp = src + len - n;

    strcpy(ssrc, tmp);
    strncat(ssrc, src, len - n);

     return true;

}

（7）、新邮通：字符串反转：给定字符串“we;tonight;you;”，编码实现输出”ew;thginot;uoy;“

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void ReverseStr( char *src)
{
     int len = strlen(src);
     int i = 0;
     int first = 0;
     int end = 0;
     while (i < len)
    {
         if (src[i] == ';')
        {
            end = i - 1;
             while (first < end)
            {
                 char tmp = src[first];
                src[first] = src[end];
                src[end] = tmp;
                first++;
                end--;
            }
            first = i + 1;

        }
        i++;
    }
}

如果给定字符串末尾没有';'，只需要修改9，10，11行

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