C++ 浮点数据到uint8_t和memcpy，返回null

我有一个浮点变量，它需要作为uint8\u t传递给函数

如何转换回浮动原始值

代码：

我试图从uint8_t数组中获取浮点原始值，需要分配给probe[channel]

uint8_t channel = message->GetOutMessage()->GetData()->data[0];
unsigned char* value = &message->GetOutMessage()->GetData()->data[1];
memcpy(&__output.output.probe[channel].dac, value, sizeof(float));

我总是得到0的值

---

请帮助

您的代码中有未定义的行为。一个

float

通常是四个字节长，而

uint8\t

是一个单字节。因此，至少要用三个字节覆盖数组

然后，当您在

GetLoopOutMessage

中传递它时，您只传递实际数组（两个字节），因此浮点数字现在丢失了它的大部分数据

您可能想使用例如

uint8_t args[1 + sizeof(float)];

---

如果

GetLoopOutMessage

没有复制您传递给它的数据，那么您还有另一种未定义的行为，即传递指向局部变量的指针。当

SetAnalog

函数返回时，存储局部变量的内存空间将被下一次函数调用重用。这意味着

args

数组不再存在于内存中，它以前所在的位置现在可能有其他数据

要解决此问题，您应该使

args

数组

静态

，使其全局化（确实不推荐），或者在堆上动态分配它：

uint8_t* args = new uint8_t[1 + sizeof(float)];

一旦您使用完此指针（即检索数据时），请不要忘记将其删除。

尝试如下操作：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

然后像这样传递：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

然后像这样读出数据：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

请注意，我在这里使用的是

new

/

delete

，而不是本地堆栈存储，因为您似乎正在回调中使用数据，所以我猜您的数据在读取之前已经被覆盖（碰巧为零）

---

如果

new

/

delete

不合适（

GetLoopOutMessage

复制数据），则按如下方式执行：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

然后像这样传递：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

然后像这样读出数据：

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = new MyArgs();
args->channel = number;
args->voltage = voltage;

....GetLoopOutMessage(....., (uint8_t *)args, sizeof(*args), ....)

struct MyArgs {
  uint8_t channel;
  float voltage;
} *args = (struct MyArgs *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

delete args;

struct {
  uint8_t channel;
  float voltage;
} args = {number, voltage};

....GetLoopOutMessage(....., (uint8_t *)&args, sizeof(args), ....)

struct mydata {
  uint8_t channel;
  float voltage;
} *args = (struct mydata *)message->GetOutMessage()->GetData()->data;

uint8_t channel = args->channel;
__output.output.probe[channel].dac = args->voltage;

---

非常感谢，但尽管更改为args[5]，该值仍然保持为0。我从未见过任何“DAC”硬件将浮点作为参数，直接或作为字节复制到数组中。args数组保持args[5]={0,0,0,64,65}，当探针号为0，电压为12时。@user2746930在快速测试我自己之后，

float

number

12.0

似乎是字节序列

0 64 65

，因此数据是正确的。然而，这可能是另一个问题，实际上是另一种未定义行为的情况。请看我的最新答案。哦，等等…'DACVoltage'-GetLoopOutMessage（）在该数组中到底需要什么？？虚拟PTCLoopOutMessage*GetLoopOutMessage（PTCHardware*源、uint8\u t地址、uint8\u t命令、const char*commandString、const uint8\u t*数据、大小\u t数据长度、PTCMessageType messageType、PTCMessagePriority）=0；这可能是可行的，但如果它希望浮点数正好在uint8之后，那么填充将是一个问题，除非结构已打包。@RetiredInja：给出的示例显示了“管道”的两端，因此我怀疑是否真的需要打包。只要他们能够更改两侧。我想

GetLoopOutMessage

可能有点像一个黑匣子。@RetiredInja:是的，它看起来像是一个直接传递数据的黑匣子。仔细想想，

sizeof

属性表明它复制了数据，因此malloc/free可能是一个糟糕的计划。