C++ Passing Struct Address

Question

Im a little bit confused about passing structs into functions. I understand pointers and everything.

But for instance:

struct stuff
{
   int one
   int two 
};

int main{
    stuff fnc;
    fnc.two = 2;
    fnc.one = 1;
    multiply(&fnc);

}

void multiply(const stuff * pm){
    cout << pm->one * pm->two;
}

First of all....am i even doing this right. And second of all, why do we use the address operator when we pass the function, but use the * pointer operator in the actual function call? Im confused?

Solution 1

Yes, your code is compilable other than the missing semicolons in the defintion of struct stuff. I'm not quite sure exactly what you're asking about passing the function and the actual function call, but I think you're wondering why the function call uses &fnc, but the parameter is stuff *pm? In that case, the fnc variable declared is a plain stuff. It is not a pointer, it refers to the actual instance of that struct.

Now the multiply function is declared as taking a stuff* -- a pointer to a stuff. This means that you can't pass fnc directly -- it's a stuff and multiply expects a *stuff. However, you can pass fnc as a stuff* by using the & operator to take the address, and &fnc is a valid stuff* that can be passed to multiply.

Once you're in the multiply function, you now have a stuff* called pm. To get the one and two variables from this stuff*, you use the pointer to member operator (->) since they are pointers to a stuff and not a plain stuff. After obtaining those values (pm->one and pm->two), the code then multiples them together before printing them out (pm->one * pm->two).

Solution 2

The * and & operands mean different things depending on whether they describe the type or describe the variable:

int  x;        // x is an integer
int* y = &x;   // y is a pointer that stores the address of x
int& z =  x;   // z is a reference to x
int  a = *y;   // a in an integer whose value is the deference of y

Your pm variable is declared as a pointer, so the stuff type is modified with *. Your fnc variable is being used (namely for its address), and thus the variable itself is marked with &.

You can imagine the above examples as the following (C++ doesn't actually have these, so don't go looking for them):

int x;
pointer<int> y = addressof(x);
reference<int> z = x;
int a = dereference(y);

It the difference between describing a type and performing an operation.

Solution 3

In

void multiply(const stuff * pm){
    cout << pm->one * pm->two;
}

The stuff * pm says that pm is an address of a stuff struct.

The

&fnc

says "the address of fnc".

When a variable is declared like:

stuff *pm;

it tells us that pm should be treated like an address whose underlying type is stuff.

And if we want to get the address of a variable stuff fnc, we must use

&fnc

Comments

[Admin]

Ya thanks, it was just an example I threw together. I was just confused at the usage of * and & operators

[Admin]

Yes this is precisely what I was asking. Thanks, my wording was terribly messed up in my question lol

[San Jacinto]

I think this is the only thing that actually answers the question.