mlsb.net
当前位置:首页 >> 如何将一个unsignED long int型数字每8位提取 >>

如何将一个unsignED long int型数字每8位提取

#include#includeint main(){byte b[4]; unsigned long int a=12345;b[0] = a&0xff;b[1] = (a>>8)&0xff;b[2] = (a>>16)&0xff;b[3] = (a>>24)&0xff;return 0;}

题目int BitCount(long l)中可不是“unsigned long”啊? 如果是long,只要用scanf("%d",&x);就可以了。 #includeint BitCount(long l){int n=0; for(;l;l/=2)n+=l%2; return n;}int main(){int n; scanf("%d",&n); printf("%d的二进制中有%d个1\n...

用位运算: short int right,left,v=0x2244; right = v & 0x00ff; // 取右8位 left = v >> 8; //取左8位 printf("right=%#x, left=%#x\n",right,left); ============== 用联合体,通过共享内存: union { short j; unsigned char a[2]; }x; shor...

unsigned long类型整数,分别将其前2个字节和后2个字节作为两个unsigned long数据类型如果占4个字节的话,比如: 5 在计算机里按二进位存储即:00000000 00000000 00000000 00000101 那么把前两个字节 00000000 00000000副给一个unsigned int变量...

分给我吧? union { unsigned long b; unsigned char c[4]; }a; unsigned int d[2]; a.b = 0x12345678; d[0] = a.c[3]; d[1] = a.c[2]; printf("%x %x\n",a.c[3],a.c[2]); printf("%x %x\n",d[0],d[1]); printf("%d %d\n",d[0],d[1]);

直接强制转化就可以了

#include int main(void){ unsigned long int value = 13; printf("value: %ld\n", value); return 0;}

a[0]=SOC>>16; printf("a[0]=%u",a[0]); a[1]=SOC; printf("a[1]=%u",a[1]);

unsigned long number;//printf("%lu", number);//

/* Arduino练习作业:写两个程序,第一个用unsigned long int做变量,输出变量转为二进制后占的bit数。 第二个用signed long int做变量,同样输出二进制所占bit数。 适用0以上正整数,0需要额外加判断 */ # include int count_bits(unsigned lon...

网站首页 | 网站地图
All rights reserved Powered by www.mlsb.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com