[英]How can I toggle certain bits in an 8bit register?
Say I have bits 0-3 I want to toggle given a certain register value, how can I do that?假设我有位 0-3 我想在给定某个寄存器值的情况下切换,我该怎么做?
eg:例如:
unsigned char regVal = 0xB5; //1011 0101
// Toggle bits 0-3 (0101) to 1 without changing 4-7 where result should be 1011 1111
unsigned char result = regVal & (0x01 | 0x02 | 0x03 | 0x04);
or或者
unsigned char regVal = 0x6D; //0110 1101
// Toggle bits 4 and 7 to 1 without changing 1,2,3,5,6 where result should be 1111 1101
unsigned char result = regVal & (0x10 | 0x80);
The way I've attempted to mask above is wrong and I am not sure what operator to use to achieve this.我试图掩盖上面的方式是错误的,我不确定要使用什么运算符来实现这一点。
To set (to 1) the particular bit:设置(为 1)特定位:
regVal |= 1 << bitnum;
To reset (to 0) the particular bit:要重置(为 0)特定位:
regVal &= ~(1 << bitnum);
Te write the value to it (it can be zero or one) you need to zero it first and then set将值写入它(它可以是零或一)你需要先将它归零然后设置
regVal &= ~(1 << bitnum);
regVal |= (val << bitnum);
unsigned char regVal = 0xB5; //1011 0101
// Toggle bits 0-3 (0101) to 1 without changing 4-7 where result should be 1011 1111
regVal |= (1 << 0) | (1 << 1) | (1 << 2) | (1 << 3);
unsigned char regVal = 0x6D; //0110 1101
// Toggle bits 4 and 7 to 1 without changing 1,2,3,5,6 where result should be 1111 1101
regVal |= (1 << 4) | (1 << 7);
unsigned char regVal = 0x6D; //0110 1101
// Set bits 4 to 7 to 1010 without changing 0, 1,2,3,
regVal &= ~((1 << 4) | (1 << 5) | (1 << 6) | (1 << 7));
regVal |= (0 << 4) | (1 << 5) | (0 << 6) | (1 << 7);
Simply use bitwise XOR.只需使用按位异或。
regval ^= 0xFu;
Or if you will regval ^= MASK;
或者,如果你愿意
regval ^= MASK;
And that's it.就是这样。
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