3.11study

#include 

/*
int main()
{
  short s = 0;
  int a = 10;
  printf("%dn",sizeof(s=a+5));//2   4个字节的整形放入2个字节的short中，应为2个字节
  printf("%dn",s);//0
  return 0;
}
*/


/*
int main()
{
  int a = 0;
  //        a--> 00000000000000000000000000000000  ->原码
  //        ~a-->11111111111111111111111111111111  ->补码
  //             11111111111111111111111111111110  ->反码
  //printf打印原码    10000000000000000000000000000001  ->原码
  printf("%dn",~a);//按二进制位取反
  // - 1

  return 0;

}
*/


/*
int main()
{
  int a = 10;
  //printf("%dn",++a);//前置++ -->先++后使用
  printf("%dn",a++);//后置++ -->先使用后++

  return 0;
}
*/

/*
int main()
{
  int a = (int) 3.14;
  printf("%dn",a);

  return 0;
}
*/

/*
void test1(int arr[])
{
  printf("%dn",sizeof(arr));
}

void test2(char ch[])
{
  printf("%dn",sizeof(ch));
}

int main()
{
  int arr[10] = { 0 };
  char ch[10] = { 0 };
  printf("%dn",sizeof(arr));//40
  printf("%dn",sizeof(ch));//10
  test1(arr);//数组传参传递的是首元素的地址  4/8字节
  test2(ch);//

  return 0;
}
*/

/*
int main()
{
  int a = 3;//011
  int b = 6;//110
  int c = a && b;
  int d = a || b;
  printf("%dn",c);//两个同时为真才为真，结果为真
  printf("%dn",d);//两个同时为假才为假，结果为真
  return 0;
}
*/

/*
int main()
{  // i=0      a=0   b=010   c=011   d=100
  int i = 0, a = 0, b = 2, c = 3, d = 4;
  i = a++ && ++b && d++;
  //a=a=0-->为假，不进行右边计算-->b=2 c=3 d=4
  printf("a=%dn, b=%dn, c=%dn, d=%dn",a,b,c,d);

  return 0;
}

*/

/*
int main()
{  // i=0      a=0   b=010   c=011   d=100
  int i = 0, a = 1, b = 2, c = 3, d = 4;
  //i = a++ && ++b && d++; //&&遇到假则停止  2334
  i = a++ || ++b || d++;  //||运算遇到真则停止 2234
  printf("a=%dn, b=%dn, c=%dn, d=%dn", a, b, c, d);

  return 0;
}
*/

/*
int main()
{
  //int a = 0;
  //int b = 0;
  //if (a > 5)
  //  b = 3;
  //else
  //  b = -3;
  //printf("%dn",b);

  int a = 0;
  int b = 0;
  b = (a > 5 ? 3 : -3);
  printf("%dn",b);

  int a = 10;
  int b = 20;
  int max = 0;
  max = (a > b ? a : b);
  printf("%dn", max);
  return 0;
}
*/

/*
int main()
{
  int a = 1;
  int b = 2;
  int c = (a > b, a = b + 10, a , b = a + 1);//ca
  printf("%d  %d  %d  n", a, b, c);

  return 0;
}
*/

/*
int main()
{
  int a[10] =
 { 0 };
  a[4] = 10;
  printf("%d  %dn",a[4],a[5]);
  return 0;
}
*/

/*
struct student
{
  char name[20];
  int age;
  char id[20];
};

int main()
{
  int a = 10;
  //使用struct student这个类型创建了一个学生对象s1，并初始化
  struct student s1 = { "张三" , 20 , "20230311" };
  struct student *ps = &s1;
  printf("%sn",(*ps).name);
  printf("%dn",(*ps).age);
  printf("%sn",(*ps).id);

  printf("%sn",ps->name);
  printf("%dn",ps->age);
  printf("%sn",ps->id );

  printf("%sn",s1.name);
  printf("%dn",s1.age);
  printf("%sn",s1.id);

  return 0;
}
*/

/*
int main()
{ //二进制数加减用补码运算
  char a = 3;//正常来说是00000000000000000000000000000011
  //char中只能存放8个字节故为00000011 -a
  char b = 127;//00000000000000000000000001111111
  //01111111 -b
  //a+b
  //    00000000000000000000000000000011
  //    00000000000000000000000001111111
  //结果  00000000000000000000000010000010
  char c = a + b;//c->10000010
  //将c按符号位补全,符号位为1
  //11111111111111111111111110000010  --补码
  //11111111111111111111111110000001  --反码
  //10000000000000000000000001111110  --原码
  //-126
  printf("%dn",c);
  return 0;
}
*/

/*
int main()
{
  char a = 0xb6;//101101110
  short b = 0xb600;
  int c = 0xb6000000;
  if (a == 0xb6)
    printf("a");
  if (b == 0xb600)
    printf("b");
  if (c == 0xb6000000)
    printf("c");
  return 0;
}
*/

/*
int main()
{//整形提升，c参与表达式运算，由1个字节变为4个字节
  char c = 1;
  printf("%un",sizeof(c));//1
  printf("%un", sizeof(+c));//4
  printf("%un", sizeof(!c));//1
  return 0;
}
*/

/*
int main()
{
  int a = 10;
  int b = 20;
  int c = b + a * 3;
  printf("%dn",c );


  return 0;
}
*/


/*
int main()
{ //无意义，++i无顺序导致结果不确定
  int i = 1;
  int c = (++i) + (++i) + (++i);
  printf("%dn",c);

  return 0;
}
*/