#include <stdio.h>
//int main()
//{
// int a=10;
// int b=20;
// printf("%d\n",a+b);
// printf("%d\n",a*b);
// printf("%d\n",a/b);//最近一次将一切都该回原始int b=20以及%d后面报出的是0,是因为int是整型10/20为0.5因此它只取前面的0
//前一次运算的时候“int b=20”而不是20.0。改成20.0还是不行。无论我d改为f,还是改回来都不行以及a=10.0我也试过了不行
//int b改为float b也不行,其他都不改变将b改为b.0会报错
// printf("%d\n",b/a);
// printf("%d\n",4+234);
// printf("%d\n",678-345);
// printf("%d\n",34/17);
// printf("%.8f\n",10/3.0);
//这里前一次在%后面输的是“8”而不是".8"导致最后输出的数为3.333333是小数点后6位而不是我设定的后8位。
// return 0;
//}
//求模运算符% /得商 %得余(仅用于整数间)
//int main()
//{
// printf("%d\n",123%10);// 123/10=12.3 最后输出3
// printf("%d\n",123%100);// 123/100=1.23 最后输出23
// printf("%d\n",1234%100);// 1234//100=12.34 最后输出34
// return 0;
//}
//负数求余:结果的正负号由第一个操作符决定
//int main()
//{
// printf("%d\n",11%-5);//输出1
// printf("%d\n",-11%5);//输出-1
// printf("%d\n",11%5);//输出1
// printf("%d\n",-11%-5);//输出-1
// return 0;
//}
//赋值操作符
//number 1.复合操作符
//int main()
//{
// int a=10;//(这叫初始化不叫赋值)
// printf("%d\n",a=20);//这叫赋值 最后输出20
// printf("%d\n",a=4+5);//这是将右边的值赋给左边 最后输出9
// return 0;
//}
//复合连赋的第二种情况
//int main()
//{
// int a=10;
// a=a+3;//或者写a+=3(这里表示一个意思:a自增三)
// printf("%d\n",a);//结果为13
// return 0;
//}
//第二种
//int main()
//{
// int a=30;
// a+=23;//(a原本的值为30,自增23后输出53)
// printf("%d\n",a);
// return 0;
//}
//第三种
//int main()
//{
// int a=10;
// a = a%3;
// printf("%.3f",a);//因为前面的a是int类型的,而这里的用了float类型,所以输出的是0.000。可以用一下强行转型的方法
// return 0;
//}
//{
// int a=10;
// a = a%3;
// printf("%.3f\n",(double)a);//因为10/3的余数为1,所以输出1(%输出余数)接下来用/试试
// return 0;
//}
//{
// int a=10;
// a =a/3;
// printf("%d\n",a);//输出整数3
// return 0;
//}
//number 2.连续赋值
//int main()
//{
// int a=3;
// int b=5;
// int c=0;
// printf("%d\n",c=b=a+3);//这里是向右赋值,即:a+3=6,那么b=6,然后c=b,即c=6
// return 0;
//}
//第二种写法,会比第一种更方便看,但不为连赋
//int main()
//{
// int a=3;
// int b=5;
// int c=0;
// printf("%d\n",b=a+3);
// printf("%d\n",c=b);//此次及上面65行均不为连赋
// return 0;
//}
//sizeof操作符
int main()
{
printf("%zu\n",sizeof(_Bool
)); printf("%zu\n",sizeof(float)); printf("%zu\n",sizeof(long long)); printf("%zu\n",sizeof(double)); printf("%zu\n",sizeof(long double)); return 0;
}
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