C语言实现strlen，atoi, strcpy等函数

strlen的实现很简单：

如下:

int my_strlen(const char* p){assert(p != NULL);int len =0;while (*p++){len++;}return len;}

 

strcmp的实现:

 

int my_cstrcmp(const char* p , const char* q){assert(p != NULL && q != NULL);//循环找到第一个不相等的字符for (; *p == * q; ++p, ++q){if (*p == 0){return 0;}}return *(unsigned char*)p - *(unsigned char*)q >0 ? 1:-1; }

 

strcpy函数实现： char* myStrcpy(char* des, const char* source){ assert(des != NULL && source != NULL); char* ret = des; while ( (*des++ = *source++) != '/0'); return ret;}  

 

itoa 函数实现：

void itoaTest(int num,char str[] ) { int sign = num,i = 0,j = 0; char temp[11]; if(sign<0)//判断是否是一个负数 { num = -num; }; do { temp[i] = num%10+'0'; num/=10; i++; }while(num>0); if(sign<0) { temp[i++] = '-'; } temp[i] = '/0'; i--; while(i>=0) { str[j] = temp[i]; j++; i--; } str[j] = '/0'; }

 

atoi函数实现：

 int atoiTest(char s[]) { int i = 0,sum = 0,sign; //输入的数前面可能还有空格或制表符应加判断 while(' '==s[i]||'/t'==s[i]) { i++; } sign = ('-'==s[i])?-1:1; if('-'==s[i]||'+'==s[i]) { i++; } while(s[i]!='/0') { if (s[i] >= 0 && s[i] <= '9') { sum = s[i]-'0'+sum*10; i++; } else { break; } } return sign*sum; }   

 反转字符串递归实现:

 char* reverse(char *str){assert(str != NULL);int iLen = strlen(str);char c;int i;if (iLen<2){return NULL;}else{//先把首尾交换c = str[0];str[0] = str[iLen-1];str[iLen-1] = c;//如果长度为2的话，返回nullif (iLen == 2){return NULL;}// 辅助空间，用来保存原字符串没有反转的部分char *str2 = (char*)malloc(iLen-1);memset(str2, '/0', iLen-1);//拷贝字符串for (i=0; i<iLen-2; i++){str2[i] = str[i+1];}//反转中间字符串reverse(str2); //把反转后的字符串copy到原字符串for (i=0; i<iLen-2; i++){str[i+1] = str2[i];}free(str2);str2 = NULL;}return str; }