将 char* 转换为 uint8_t
我通过 CAN 协议传输消息.
为此，CAN 消息需要 uint8_t 类型的数据.所以我需要将我的 char* 转换为 uint8_t.通过对本网站的研究，我生成了以下代码:
To do so, the CAN message needs data of uint8_t type. So I need to convert my char* to uint8_t. With my research on this site, I produce this code :
char* bufferSlidePressure = ui->canDataModifiableTableWidget->item(6,3)->text().toUtf8().data();//My char*
/* Conversion */
uint8_t slidePressure [8];
sscanf(bufferSlidePressure,"%c",
&slidePressure[0]);
如您所见，我的 char* 必须适合 sliderPressure[0].
As you may see, my char* must fit in sliderPressure[0].
我的问题是即使我在编译过程中没有错误，slidePressure中的数据完全不正确.确实，我用 char* = 0 对其进行了测试，并且我有未知字符......所以我认为问题必须来自转换.
My problem is that even if I have no error during compilation, the data in slidePressure are totally incorrect. Indeed, I test it with a char* = 0 and I 've got unknow characters ... So I think the problem must come from conversion.
我的数据可以是Bool、Uchar、Ushort和float.
感谢您的帮助.
推荐答案
你的字符串是整数吗?例如.char* bufferSlidePressure = "123";?
Is your string an integer? E.g. char* bufferSlidePressure = "123";?
如果是这样，我会这样做:
If so, I would simply do:
uint8_t slidePressure = (uint8_t)atoi(bufferSlidePressure);
或者，如果您需要将其放入数组中:
Or, if you need to put it in an array:
slidePressure[0] = (uint8_t)atoi(bufferSlidePressure);
根据您的评论，如果您的数据可以是任何内容，我想您必须将其复制到新数据类型的缓冲区中.例如.类似:
Following your comment, if your data could be anything, I guess you would have to copy it into the buffer of the new data type. E.g. something like:
/* in case you'd expect a float*/
float slidePressure;
memcpy(&slidePressure, bufferSlidePressure, sizeof(float));
/* in case you'd expect a bool*/
bool isSlidePressure;
memcpy(&isSlidePressure, bufferSlidePressure, sizeof(bool));
/*same thing for uint8_t, etc */
/* in case you'd expect char buffer, just a byte to byte copy */
char * slidePressure = new char[ size ]; // or a stack buffer
memcpy(slidePressure, (const char*)bufferSlidePressure, size ); // no sizeof, since sizeof(char)=1
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