作为参数传递时，在函数中更新指针地址。 [将指针作为C中的引用传递]

Question

CONCEPT: Passing pointer by reference

Trying to achieve: To get updated pointer address from function, when passed as an argument.

int main(void)
{
    uint8_t unArray[10] = {0};  // unint8_t is type def as unsigned char
    uint8_t * pmyPtr;

    pmyPtr = unArray;

    func(pmyPtr);

    *pmyPtr = someValue3; 

}

void func(uint8_t * unPtr)
{
    *unPtr = someValue1;
     unPtr++;
    *unPtr = someValue2;  
     unPtr++;  

}

Suppose we have unArray address as 0x0001000. So pmyPtr will have 0x0001000 as its assigned a constant pointer.

When pointer is passed to the function func some indexes of array (first two) are updated by DE-referencing.

When I come back to the main after func execution I am trying to update the third index. How can this be achieved. I have a hunch that double De-referencing might be handy.

Answer 1

You are correct. A pointer to pointer is a simple solution. In c++ it could be syntactically hidden as a reference.

main()
{
    uint8_t unArray[10] = {0};  // unint8_t is type def as unsigned char
    uint8_t * myPtr;

    myPtr = unArray;

    func(&myPtr);//NOTICE ADDRESS TAKING

    *myPtr = someValue3; 

}

void func(uint8_t ** unPtrPtr)//ONE MORE STAR
{
    uint8_t * unPtr=*unPtrPtr;//CHANGED
    *unPtr = someValue1;
     unPtr++;
    *unPtr = someValue2;  
     unPtr++;
    *unPtrPtr = unPtr;//CHANGED
}