3.15 3.16study

#include <stdio.h>

/*
int main()
{//数组内容交换
  int arr1[] = { 1,3,5,7,9 };
  int arr2[] = { 2,4,6,8,0 };

  int tmp = 0;
  int i = 0;
  int sz = sizeof(arr1) / sizeof(arr1[0]);
  for (i = 0; i < sz; i++)
  {
    tmp = arr1[i];
    arr1[i] = arr2[i];
    arr2[i] = tmp;
    printf("%d arr1:%d arr2:%d\n", i, arr1[i], arr2[i]);
  }

  return 0;
}
*/

/*
int main()
{
  int arr[] = {1,2,3,4,5};//int*是4个字节
  short* p = (short*)arr;//short*只能访问两个字节
  int i = 0;
  for (i = 0; i < 4; i++)
  {
    *(p + i) = 0;//仅修改了1-->0000  2-->0002
  }
  for (i = 0; i < 5; i++)
  {
    printf("%d ",arr[i]);//00345
  }

  return 0;
}
*/

/*
int main()
{
  int a = 0x11223344;//存储为 44 33 22 11，两个字节存放一起
  char* pc = (char*)&a;
  *pc = 0;//仅能操作一个字节即内存为00 33 22 11
  printf("%x\n",a); //11223300
   

  return 0;
}
*/

/*
int i;//全局变量--若不初始化，默认为0
int main()
{
  i--;//i=-1 -->10000000000000000000000000000001原码
  //11111111111111111111111111111110反码
  //11111111111111111111111111111111补码-->很大的数>4-->sizeof(i)
  if (i > sizeof(i))//sizeof--计算变量/类型所占内存的大小 >=0 无符号数
  {
    printf(">\n");
  }
  else
  {
    printf("<\n");
  }
  //>

  return 0;
}
*/

/*
int main()
{
  int a, b, c;
  a = 5;
  c = ++a;//先++后赋值给c，即a=a+1,c=a
  //c=6 a=6
  b = ++c, c++, ++a, a++;//c=c+1,c=c+1,a=a+1,b=a++-->b=a,a=a+1
  //c=8,a=8,b=7
  //逗号表达式，从左到右计算，最后一个逗号内的赋值给等号前的
  b += a++ + c;//b=b+a+c,a++
  //+的优先级高于++  ,a=9 c=8,b=23
  printf(" a=%d\n b=%d\n c=%d\n",a,b,c);
  return 0;
}
*/

/*
int count_bit_one(int a)
{
  int count = 0;
  while (a)
  {
    if (a % 2 == 1)
    {
      count++;
    }
    a = a / 2;
  }
  return count;
}
//只适用于正数
int main()
{//求a的2进制表示中有几个1
  int a = 0;//e.g. 13-->0000000000000000000000001101
  scanf("%d",&a);
  int count = count_bit_one(a);
  printf("count= %d\n",count);
  return 0;

}
*/

/*
int count_bit_one(unsigned int a)//使用无符号数unsigned int 避免负数影响结果
{
  int count = 0;
  while (a)
  {
    if (a % 2 == 1)
    {
      count++;
    }
    a = a / 2;
  }
  return count;
}
//只适用于正数
int main()
{//求a的2进制表示中有几个1
  int a = 0;//e.g. 13-->0000000000000000000000001101
  scanf("%d", &a);
  int count = count_bit_one(a);
  printf("count= %d\n", count);
  return 0;

}
*/

/*
int count_bit_one(int a)//使用无符号数unsigned int 避免负数影响结果
{
  int count = 0;
  int i = 0;
  for(i = 0; i < 32; i++)
  {
    if ((a >> i) & 1 == 1)
      count++;
  }
  return count;
}
//只适用于正数
int main()
{//求a的2进制表示中有几个1
  int a = 0;//e.g. 13-->0000000000000000000000001101
  scanf("%d", &a);
  int count = count_bit_one(a);
  printf("count= %d\n", count);
  return 0;

}
*/

/*
int count_bit_one(int a)//使用无符号数unsigned int 避免负数影响结果
{
  int count = 0;
  while (a)
  {
    a = a & (a - 1);//每进行1一次消去一个1
    //a=13    1101
    //a-1   1100
    //新a    1100    count=1
    //新a-1    1011
    //新新a   1000    count=2
    //新新a-1 0111
    //新新新a  0000    count=3
    count++;
  }
  return count;
}
//只适用于正数
int main()
{//求a的2进制表示中有几个1
  int a = 0;//e.g. 13-->0000000000000000000000001101
  scanf("%d", &a);
  int count = count_bit_one(a);
  printf("count= %d\n", count);
  return 0;

}
*/

/*
int get_diff_bit(int a, int b)
{
  int tmp= a^b;
  int count = 0;
  while (tmp)
  {
    tmp = tmp & (tmp - 1);
      count++;
  }
  return count;
}

int main()
{//求二进制中不同位的个数
  int a = 0;
  int b = 0;
  scanf("%d%d", &a, &b);
  int count = get_diff_bit(a, b);

  printf("%d\n",count);
  return 0;
}
*/