如何正确地设置指向函数的指针

试试这个：CANRxMessageBuffer*Rxmessage;Rxmessage=(CANRxMessageBuffer*)CANGetRxMessage(CAN_USED,CAN_RX_CHANNEL);//读取缓冲区



      以下为原文

    Try this:

CANRxMessageBuffer * Rxmessage;
Rxmessage = (CANRxMessageBuffer * ) CANGetRxMessage(CAN_USED,CAN_RX_CHANNEL);//Read the buffer

嗨，当然，*创建指针。在XC32中，CANGetRxMessage给了我们一个指针，我没有选择。所以我需要将这个指针传递到我的函数。



      以下为原文

    Hi,
of course  the * create the pointer.
in XC32 the CANGetRxMessage give us a pointer I've no choice.
so I need to pass this pointer to my function.

Hi nsSuple但RxMead仍然是指针



      以下为原文

    Hi N_S_pole
But Rxmessage is still a pointer with this

我试着从逻辑上说它不起作用。



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    I try and logically it don't works
 

你把RxMead指向缓冲区了吗？或者CangeTrxMaseAGE（）应该这样做？



      以下为原文

    Have you pointed Rxmessage to a buffer?  Or is CANGetRxMessage() supposed to do that?

指针很好。问题是指针指向的缓冲区在哪里？



      以下为原文

    A pointer is fine.
The question is where is the buffer the pointer is pointing too?

指针声明为thisCANRxMessageBuffer*Rxmessage；CANRxMessageBuffer声明为thistypedef union{struct{//这是CAN RX消息的SID部分。CAN_RX_MSG_SID msgSID；//这是CAN RX消息CAN_MSG_EID msgEID的EID部分；//这是//接收到的消息的数据有效负载部分。字节数据[8 ]；}；//如果消息缓冲器是/或从一个数据类型的CAN RX信道读取的话，这是可以使用的。字节DATAONLYSLGDATA（8）；//这是可以作为32位/ /字组组织的RX消息。UINT32 messageWord[4];}CANRxMessageBuffer；函数CANGetRxMessage在编译器CANRxMessageBuffer*CANGetRxMessage（CAN_MODULE模块，CAN_CHANNEL通道）中这样声明；



      以下为原文

    The pointer is declared like this
CANRxMessageBuffer * Rxmessage;
and CANRxMessageBuffer is declared like this
typedef union {

    struct
    {
        // This is SID portion of the CAN RX message.
        CAN_RX_MSG_SID msgSID;

        // This is EID portion of the CAN RX message
        CAN_MSG_EID msgEID;

        // This is the data payload section of the
        // received message.
        BYTE data[8];
    };

    // This can be used if the message buffer is to
    // be read from a Data-Only type of CAN RX Channel.
    BYTE dataOnlyMsgData[8];

    // This is CAN RX message organized as a set of 32 bit
    // words.
    UINT32 messageWord[4];

}CANRxMessageBuffer;
 
the function CANGetRxMessage is declared like this in the compiler
 
CANRxMessageBuffer * CANGetRxMessage(CAN_MODULE module, CAN_CHANNEL channel);

你知道数据指针和数据本身的区别吗？C不自动分配数据。作为一个程序员，你必须做到这一点。如果你有一个指向数据的指针。然后直接传递指针。没有星星。（*）如果您使用的数据是AN＆AMP；它的意思是“地址”



      以下为原文

    Do you understand the difference between a pointer to data and the data itself?
C does not automatically allocate data.
You as a programer must do it.
If you have a pointer that points at data. Then pass the pointer directly. No stars. (*)
If you have the data you use an &.
It means "the address of"

没错，它仍然有效。我不确定从Chanz收到的返回类型。你说你认为它是指针，但不确定。你可能会遇到一个例外。如果它是一个例外，为什么不检查调用栈中的原因的确切位置。



      以下为原文

    That is true it should still work. I was not sure the return type from the Can_receive.
 
You said you think its a pointer but not sure. You might be getting an exception. if its an exception why dont you check the exact location of the cause in the Call stack.

只是猜测，如果没有新消息，CangeTrxMeXAGE（）返回null，需要处理吗？



      以下为原文

    Just a guess, might CANGetRxMessage() return NULL if there is no new message, and you need to handle that?

所以，是的，你需要检查NULL。



      以下为原文

   
/***************************

 * CAN Receive

 * *************************/
CANRxMessageBuffer * CANGetRxMessage(CAN_MODULE module, CAN_CHANNEL channel)
{
    /* This function return a pointer to
     * the next message to read.
     */
    CAN_REGISTERS * canRegisters = (CAN_REGISTERS *) canModules[module];

    if (canRegisters->canFifoRegisters[channel].CxFIFOINTbits.RXNEMPTYIF == 0)
    {
        return (NULL);
    }
    else
    {
        return ((CANRxMessageBuffer *) PA_TO_KVA1(canRegisters->canFifoRegisters[channel].CxFIFOUA));
    }
}

So, yes, you need to check for NULL.