[英]How do I Store 2 8bit Register data into 1 16bit varrible?
I am looking at taking the 10 bit data from my ADC conversion and storing it into 1 16 bit integer data looks like 0x03 ADRESH 0xFF ADRESL. 我正在考虑从ADC转换中获取10位数据并将其存储为1个16位整数数据,如0x03 ADRESH 0xFF ADRESL。 What I am doing right now is
我现在正在做的是
data = 0x03 & ADRESH;
data = data << 8;
data = data & 0x03FF & ADRESL;
will this work how I think it should or am I missing something? 这项工作会以我认为应该的方式进行还是我会丢失一些东西? thanks for the help
谢谢您的帮助
Why don't you use the |
你为什么不用
|
operator ? 操作员?
short data = ((0x03 & ADRESH) << 8) | ADRESL;
should work fine. 应该工作正常。
Your code will not work 你的代码不起作用
data = data & 0x03FF & ADRESL;
Should be closer to 应该更接近
data = data | ADRESL;
or
data |= ADRESL;
It is good that you performed the 8 byte shift in your 16-bit data
. 在16位
data
执行8字节移位是很好的。
Note: the & 0x03FF
is not needed. 注意:不需要
& 0x03FF
。
Note: Insure the data type of data
is at least 16 bits. 注意:确保的数据类型
data
是至少16位。
Note: If you continue to have issues, insure the 10 bit alignment is as you think. 注意:如果您仍然遇到问题,请确保10位对齐符合您的想法。 Many A/D modules allow the 10-bit data to be in the upper 10 bits rather than the lower.
许多A / D模块允许10位数据位于高10位而不是低位。
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