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匿名用户
1级
2012-07-19 回答
按照你写的这个函数功能:(下边为演示运算过程)
unsigned int DatTemp = 0x12345678; (十六进制的)
unsigned int a,b,c,temp;
temp = DatTemp &0xff; 此时temp = 0x78 (十六进制的)
假设:
a = temp&0xf; //此处a = 8 ,我想愿意你应该写成 a = temp&0xf0 这样就可以保存数字7
b = a>>4; //此处b 是由 a=8 右移4位得到的,故而b = 0
c = temp&0xf; // c = 8
那么 原来的最后一句表达式 temp = ((temp&0xf)>>4)*10+temp&0xf;可以转化为
temp = b *10 + c ;因此实际上只有最末4bit才被保存下来了,最终temp的结果为8
楼主说的保留最低两位 是指 DatTemp = 0x12345678; 的最后两个数字78吧,至于得到十进制数,不知道你是想怎样得到
第一种是 将最后两个数字直接由十六进制转换得到十进制
temp = ((temp&0xf0)>>4)*16+temp&0x0f; //这样就相当于 7*16 + 8
第二种就是楼主上边写的:但是必须注意 是 &0xf0
temp = ((temp&0xf0)>>4)*10+temp&0x0f; //这里就相当于是 7*10 + 8
之所以你的只运行了最后4bit,是因为你在计算保留上边的倒数第二个数字的时候
应该写成 temp & 0xf0 ,而不是你写的temp &0xf
在计算机中,最低两位 与 最后的两位数是有很大区别的;最低两位多半是指二进制形式下的最低两位,如果是那样的话,要想保存最低两位
temp = DatTemp & 00000011(也可以写成 temp = DatTemp & 0x3)
追问:
那个&0x0f的确是写错咯,应该位0XF0,我本意也是如此,但结果任然不对哈,temp =
((temp&0xf0)>>4)*10+temp&0x0f;这个语句得不到78,我想知道这个是为什么?
追答:
那是因为 你后边 temp&0x0f 没有加括号: 应该写成 temp = ((temp&0xf0)>>4)*10+ ( temp&0x0f ); 因为 +
运算符的优先级高于 & 运算符;你上边那样写,是先 70 + 0x78 & 0x0f = (77+0x78) & 0x0f = 0xbe & 0x0f =
0x0e (打印十进制为 14)