加入十六进制值来创建一个16位值 [英] Joining hex values to create a 16bit value

问题描述

林收到这是最，中东和三个UINT8值至少地块价值的有效位数：

Im receiving three uint8 values which are the Most, Middle and Least Significant Digits of a plot value:

EG：在控制台打印（％C）：

EG: Printed in console (%c):

   1 A 4

我需要将它们传递到它接受uint16_t的信号视图界面图示。到目前为止，即时通讯做的方式工作不正常。

I need to pass them into a signal view UI grapher which accepts a uint16_t. So far the way im doing it is not working correctly.

 uint16_t iChanI = (bgp->iChanIH << 8) + (bgp->iChanIM <<4 ) + bgp->iChanIL;
 uint16_t iChanQ = (bgp->iChanQH << 8) + (bgp->iChanQM <<4) + bgp->iChanQL;

[self updateSView:iChanI ichanQ:iChanQ];

难道我正确地将它们合并，或只是增加值？

Am i merging them correctly, or just adding the values?

任何帮助很多AP preciated，

Any help is much appreciated,

谢谢，

推荐答案

您首先需要每一个十六进制字符转换为其对应的4位（半字节）再presentation，然后将它们合并成一个 int16_t ，例如：

You first need to convert each hex character to its equivalent 4 bit (nybble) representation, and then merge them into an int16_t, e.g.

uint8_t to_nybble(char c)
{
    return 'c' >= '0' && c <= '9' ? c - '0' : c - 'A' + 10;
}

uint16_t iChanI = (to_nybble(bgp->iChanIH) << 8) |
                  (to_nybble(bgp->iChanIM) << 4) |
                   to_nybble(bgp->iChanIL);

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