int my_strlen(const char *str){
  assert(str);
  if(*str=='\0'){
    return 0;
  }
  return 1 + my_strlen(str + 1);//采用遞歸方式
}

char *my_strcpy(char *dest,const char *sorc)
{
  assert(dest && sorc);
  char *ret = dest;
  while (*dest++ = *sorc++)//當(dāng)/0被最后一次拷入時(shí)，跳出循環(huán)
  {
    ;
  }
  return ret;
}

char *my_strcat(char *dest, const char *sorc)
{
  char *ret = dest;
  while (*dest != '\0')
  {
    dest++;
  }
  while (*dest++ = *sorc++)
  {
    ;
  }
  return ret;
}

int my_strcmp(const char *str1, const char *str2)
{
  assert(str1 && str2);
  while (*str1 == *str2)
  {
    if (*str1 == '\0' && *str2 == '\0')
    {
      return 0;
    }
    str1++;
    str2++;
  }
  return *str1 - *str2;
}

char *my_strstr(char *mum, char *child)
{
  assert(mum && child);
  const char *pc = mum;
  while (*pc)
  {
    const char *p1 = pc;
    const char *p2 = child;
    while (*p1 == *p2 && *p1 && *p2)
    {
      p1++;
      p2++;
    }
    if (*p2 == '\0')
    {
      return (char *)pc;
    }
    pc++;
  }
  return NULL;
}

int main()
{
  char arr[]="skyline&csdn.com";
  char arr2[30]={0};//因?yàn)閟trtok函數(shù)會(huì)修改源數(shù)組，通常復(fù)制后處理
  char sep[]="&.";
  strcpy(arr2,arr);
  printf("%s\n",strtok(arr2,sep));
  printf("%s\n",strtok(NULL,sep));
  printf("%s\n",strtok(NULL,sep));
  return 0;
}

int main()
{
  prinf("%s\n",strerror(errno));//根據(jù)程序出現(xiàn)的問(wèn)題輸出報(bào)錯(cuò)字符串
}

int main()
{
  int num1[6] = {1, 2, 3, 4, 5, 6};
  my_memcpy(num1+2, num1, 16);//希望得到121234
                //實(shí)際得到121212
  for (int i = 0; i < 6;i++){
    printf("%d", num1[i]);
  }
  return 0;
}

void my_memcpy(void *dest, void *src, size_t count){
  assert(dest && src);
  while (count--)
    {
      *(((char *)dest)) = *(((char *)src));
      dest=(char*)dest+1;//最好不要寫成++(char*)dest，gcc編譯器認(rèn)為(char*)dest不可作為左值
      src=(char*)src+1;
    }
}

void my_memmove(void *dest, void *src, size_t count)
{
  assert(dest && src);
  if (dest > src)//當(dāng)目標(biāo)位置在源位置前，我們要從后向前拷貝
  {
    while (count--)
    {
      *(((char *)dest) + count) = *(((char *)src) + count);
    }
  }
  else
  {
    while (count--)//)//當(dāng)目標(biāo)位置在源位置后，我們要從前向后拷貝
    {
      *(((char *)dest)) = *(((char *)src));
      dest=(char*)dest+1;
      src=(char*)src+1;
    }
  }
}