模拟实现memcpy/memmove/memset

1.模拟实现memcpy

//memcmp
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
int my_memcmp(const char* str1, const char* str2, int num)
{
  assert(str1 != NULL &&str2 != NULL);
  while (num--&&*str1==*str2)
  {
    str1++;
    str2++;
  }
  if (num == 0)
    return 0;
  else
    return *str1 - *str2;

}
int main()
{
  char str1[] = "aabbccdd";
  char str2[] = "aabbccdf";
  int ret=my_memcmp(str1, str2, sizeof(str1));
  printf("%d\n", ret);
  system("pause");
  return 0;
}

2.模拟实现memmove

memmove 
//从源目标字符串中，拷贝count个字符，如果出现内存重叠问题，就会从后向前进行拷贝
//如果没有，就会想memcopy一样进行拷贝

#include<stdio.h>  
#include<assert.h>  

void *my_memmove(void *dest, void *str, size_t len)
{
  assert(dest);
  assert(str);
  char *pdest = (char *)dest;
  char *pstr = (char*)str;
  if ((pstr < pdest) && (pdest < pstr + len))  //源和目标有重叠的情况，从后往前拷贝  
  {
    while (len--)
    {
      *(pdest + len) = *(pstr + len);
    }
  }
  else            //没有重叠，从前往后拷贝  
  {
    while (len--)
    {
      *pdest = *pstr;
      pdest++;
      pstr++;
    }
  }
  return dest;
}

int main()
{
  int i = 0;
  int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
  int size = sizeof(arr) / sizeof(arr[0]);
  my_memmove(arr, arr + 3, 8);  //注意这里的长度是以一字节为单位的  
  for (i = 0; i <size; i++)
  {
    printf("%d ", arr[i]);
  }

  system("pause");
  return 0;
}

3.模拟实现memset

//memset
#include<stdio.h>
#include<assert.h>
void* my_memset(char * ptr, int value, size_t num)
{
  assert(ptr);
  for (int i = 0; i < num; i++)
  {
    ptr[i] = value;
  }
}
int main()
{
  char str[] = "almost every programmer should know memset!";
  my_memset(str, '-', 6);
  printf("%s\n", str);
  system("pause");
  return 0;
}