C++ 将 char* 转换为 uint8_t
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Convert char* to uint8_t
提问by Evans Belloeil
I transfer message trough a CAN protocol.
我通过 CAN 协议传输消息。
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 :
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：
char* bufferSlidePressure = ui->canDataModifiableTableWidget->item(6,3)->text().toUtf8().data();//My char*
/* Conversion */
uint8_t slidePressure [8];
sscanf(bufferSlidePressure,"%c",
&slidePressure[0]);
As you may see, my char*must fit in sliderPressure[0].
如您所见，我char*必须适合sliderPressure[0].
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* = 0and I 've got unknow characters ... So I think the problem must come from conversion.
我的问题是，即使我在编译过程中没有错误，slidePressure 中的数据也完全不正确。事实上，我用 a 测试它，char* = 0我有未知的字符......所以我认为问题必须来自转换。
My datas can be Bool, Uchar, Ushort and float.
我的数据可以Bool, Uchar, Ushort and float。
Thanks for your help.
谢谢你的帮助。
回答by carmellose
Is your string an integer? E.g. char* bufferSlidePressure = "123";?
你的字符串是整数吗？例如char* bufferSlidePressure = "123";？
If so, I would simply do:
如果是这样，我会简单地做：
uint8_t slidePressure = (uint8_t)atoi(bufferSlidePressure);
Or, if you need to put it in an array:
或者，如果你需要把它放在一个数组中：
slidePressure[0] = (uint8_t)atoi(bufferSlidePressure);
Edit: 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:
编辑：按照您的评论，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：
/* 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
回答by Andy Newman
uint8_t is 8 bits of memory, and can store values from 0 to 255
uint8_t 是 8 位内存，可以存储 0 到 255 的值
char is probably 8 bits of memory
char 可能是 8 位内存
char * is probably 32 or 64 bits of memory containing the address of a different place in memory in which there is a char
char * 可能是 32 或 64 位内存，其中包含内存中不同位置的地址，其中有一个字符
First, make sure you don't try to put the memory address (the char *) into the uint8 - put what it points to in:
首先，确保您不要尝试将内存地址（char *）放入 uint8 - 将它指向的内容放入：
char from;
char * pfrom = &from;
uint8_t to;
to = *pfrom;
Then work out what you are really trying to do ... because this isn't quite making sense. For example, a float is probably 32 or 64 bits of memory. If you think there is a float somewhere in your char * data you have a lot of explaining to do before we can help :/
然后找出你真正想做的事情……因为这不太合理。例如，浮点数可能是 32 或 64 位内存。如果您认为 char * 数据中的某处存在浮点数，那么在我们提供帮助之前，您需要进行大量解释：/
回答by CashCow
char *is a pointer, not a single character. It is possible that it points to the character you want.
char *是一个指针，而不是单个字符。它可能指向您想要的字符。
uint8_tis unsigned but on most systems will be the same size as a charand you can simply cast the value.
uint8_t是无符号的，但在大多数系统上与 a 的大小相同char，您可以简单地转换该值。
You may need to manage the memory and lifetime of what your function returns. This could be done with vector< unsigned char>as the return type of your function rather than char *, especially if toUtf8() has to create the memory for the data.
您可能需要管理函数返回的内存和生命周期。这可以vector< unsigned char>作为函数的返回类型而不是 来完成char *，特别是如果 toUtf8() 必须为数据创建内存。
Your question is totally ambiguous.
你的问题完全模棱两可。
ui->canDataModifiableTableWidget->item(6,3)->text().toUtf8().data();
That is a lot of cascading calls. We have no idea what any of them do and whether they are yours or not. It looks dangerous.
这是很多级联调用。我们不知道他们做了什么，也不知道他们是否属于你。看起来很危险。
回答by CashCow
More safe example in C++ way
C++ 中更安全的例子
char* bufferSlidePressure = "123";
std::string buffer(bufferSlidePressure);
std::stringstream stream;
stream << str;
int n = 0;
// convert to int
if (!(stream >> n)){
//could not convert
}
Also, if boost is availabe
此外，如果提升可用
int n = boost::lexical_cast<int>( str )
