将指针的地址而不是指针作为 function 参数传递

Question

So I'm trying to understand the argument passing to this function.

The second argument takes a pointer

BaseType_t xQueueGenericSend( QueueHandle_t xQueue, const void * const pvItemToQueue, TickType_t xTicksToWait, const BaseType_t xCopyPosition ) PRIVILEGED_FUNCTION;

then how come passing an address of a pointer is legit?

struct AMessage *pxMessage;
pxMessage = & xMessage;
xQueueGenericSend( xQueue2, ( void * ) &pxMessage, ( portTickType ) 0, queueSEND_TO_BACK );
 

Reference: http://web.ist.utl.pt/~ist11993/FRTOS-API/group___queue_management.html#xQueueGenericSend

Answer 1

If a subroutine wants to output a pointer, like where in the parse of your string it stopped, you pass it a pointer to that pointer, so it can write your pointer with that output.

Casting makes type checking quiet but you can ask for silly things. A void pointer, by definition, can be cast or assigned to point to anything, including a pointer, a pointer to a pointer, a pointer to a pointer to a pointer, etc. A pointer is just an unsigned long with a big attitude, the type it points to, and a cast changes that.

Answer 2

freeRTOS queue function passes the message. If the message is the pointer you need to pas reference to it. Then this reference is being dereferenced.

In this case the pointer to message is going to be queued - not the message itself.

It works exactly as memcpy. memcpy parameters are void * and any pointer can be passed to this function.