使用位

Question

计划:

typedef bitset<8> bits;
char original = 0xF0F0F0F0;
char Mask = 0xFFFF0000;
char newBits = 0x0000AAAA;

/*& operation with "0bit set 0" & "1bit give no change to original byte" */
cout<<"Original o: "<<bits(original)<<endl;
cout<<"NewBits: "<<bits(newBits)<<endl;
cout<<"Mask m: "<<bits(Mask)<<endl;
cout<<"o & m with Mask: "<<bits(original & Mask)<<endl;/*0 set original bit as 0 */

结果:

原始o:11110000
NewBits:10101010
面具m:00000000
o&m with Mask:00000000
结果10101010

我理解十六进制及其结果..但....... o & m== 0000 0000所以bits(o & m | newBits)结果应该是0000 0000,而不是1010 1010......

我错过了这个概念......

任何人都可以帮助我...

期待良好的回应

谢谢

Answer 1

o & m = 0000 0000和newBits = 1010 1010.所以,如果你或他们(位),你将得到的结果1010 1010作为0|0=0, 0|1=1, 1|0=1, 1|1=1.

0000 0000 OR WITH
1010 1010
-----------------
1010 1010
-----------------