#include<stdio.h>
int main()
{
	//取余（模）、除法
	int ret1 = 10 % 3;//取余（模），得到的結(jié)果是相除之后的余數(shù)
	int ret2 = 10 / 3;//除法-得到的結(jié)果是商
 
	printf("ret1=%d\nret2=%d\n", ret1, ret2);
 
	//當(dāng)/（除號(hào)）兩端都是整數(shù)的時(shí)候，執(zhí)行的是整數(shù)除法
	//當(dāng)兩端只要有一個(gè)浮點(diǎn)數(shù)，執(zhí)行的就是浮點(diǎn)數(shù)的除法
	double ret = 10.0 / 3;
	printf("%lf\n", ret);
 
	return 0;
}

#include<stdio.h>
int main()
{
	//2.移位操作符
	//左移
	int a = 5;
	//把a(bǔ)在內(nèi)存中存儲(chǔ)的二進(jìn)制位向左移動(dòng)2位
	int b = a << 2;
 
	printf("%d\n", b);
	printf("%d\n", a);
}

#include<stdio.h>
int main()
{
	int a = 5;
	int b = a >> 2;
	printf("%d\n", b);
    return 0;
}

#include<stdio.h>
int main()
{
    int a = 3;
	int b = -5;
	int x = a & b;
	int y = a | b;
	int z = a ^ b;
	printf("%d %d %d", x, y, z);
 
	return 0;
}

#include<stdio.h>
int main()
{
	int a = 3, b = 5;
	printf("交換前：a=%d b=%d\n", a, b);
	a = a + b;//a+b=8
	b = a - b;//此時(shí)的a=8,b還是5，則b=8-5=3，也就是之前的a
	a = a - b;//次時(shí)的a還是8，而b已經(jīng)賦值為3，所以這時(shí)的a=8-3=5，也就是之前的b
	printf("交換后：a=%d b=%d\n", a, b);
 
	return 0;
}

#include<stdio.h>
int main()
{
	int a = 3, b = 5;
	printf("交換前：a=%d b=%d\n", a, b);
	a = a ^ b;
	b = a ^ b;//相當(dāng)于a^b^b=a
	a = a ^ b;//相當(dāng)于a^b^a（此時(shí)的b=a）=b
	printf("交換后：a=%d b=%d\n", a, b);
 
	return 0;
}