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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一、实现字符串操作标准库函数

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

// 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;} SQL Code 123456789101112131415161718/* 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 = (char*)malloc(sizeof(char)*n);    size_t i;    char *d = (char*)dest;    const char *s = (const char*)src;    for (i = 0; i < n; i++)        temp[i] = s[i];    for (i = 0; i < n; i++)        d[i] = temp[i];    free(temp);    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;}

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

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;}

不用临时空间的memmove实现：

//src可以不保留void *memmove(void *dst, const void *src, size_t count){    byte *pbTo = (byte *)dst;    byte *pbFrom = (byte *)src;    assert(dst != NULL && src != NULL);//不能存在空指针    if (dst <= src || pbTo >= pbFrom + count)//    {        while (count-- > 0)        {            *pbTo++ = *pbFrom++; //按递增拷贝        }    }    else  //    {        pbTo = pbTo + count - 1; //overlap的情况，从高位地址向低位拷贝        pbFrom = pbFrom + count - 1;        while (count-- > 0)        {            *pbTo-- = *pbFrom--; //按递减拷贝        }    }    return dst;}

memset的实现:

void *memset(void *buffer, int c, int count){    char *buffer_p = (char *)buffer;    assert(buffer != NULL);    while(count-- > 0)        *buffer_p++ = (char)c;    return buffer;}

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。

//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。

/*************************************************************************    > 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 10typedef 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movl 指令把字符串写到栈上,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

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

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

// 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”。

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”。

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;}

更巧妙的方法：

void reverse(char* str, int left, int right){    char tmp;    while (left < right)    {        tmp = str[left];        str[left++] = str[right];        str[right--] = tmp;    }}void rightmove(char* str, int k){    n = strlen(str);    k = k % n;    reverse(str, 0, n-k-1);    reverse(str, n-k, n-1);    reverse(str, 0, n-1);}

（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;“

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

if (src[i] == ';' || i == len - 1){    if (src[i] == ';')        end = i - 1;    else        end = i;    while...}

（8）、不使用局部变量实现strlen、两数交换

#define swap(a,b) \{ assert(sizeof(a)==sizeof(b)); char tempBuf[sizeof(a)]; memcpy(tempBuf,&a,sizeof(a)); memcpy(&a,&b,sizeof(b)); memcpy(&b,tempBuf,sizeof(b)); }#define swap(a, b) \    do { typeof(a) __tmp = (a); (a) = (b); (b) = __tmp; } while (0)// typeof 是gcc支持，iso c支持__typeof__int mstrlen(char *p){    return ToEnd(p)-p;}char * ToEnd(char * p){    while(*p != '\0')        p++;    return p;}

转载自http://blog.csdn.net/jnu_simba/article/details/11714251

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