实现一些字符串操作标准库函数、解决一些字符串问题

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