Passing Pointers to Functions

[Sw07fSEw]

So for background, I've been messing around doing examples on my own trying to further my understanding of pointers. So far I think I've been progressing but I can't wrap my head around one concept. I think my train of thought, logic, or something is wrong but I don't understand why. I'll provide two examples of code below to illustrate what I'm talking about.

Example 1:

  void adjustMinWage(double *p_minWage);

int main(){
	double minWage = 8.75;
	
	adjustMinWage(&minWage); //were passing the address of minWage
	cout << "The minimum wage has been adjusted to: $" << minWage;
	
	return 0;
}

void adjustMinWage(double *p_minWage){ //creates a local pointer, p_minWage and assigns it the address from &minWage
	cout << "What would you like to change the new minimum wage to? $";
	cin >> *p_minWage; // The value at the address of p_minWage (the same address as &minWage) is adjusted to some user input.
}

I don't really have any questions on the above example, I think my comments explain the procedure, if they do not, please correct me. The example above to more so a reference for the example below.

void adjustMinWage(double &minWage); //what does this function take?

int main(){
	double minWage = 8.75;
	double *p_minWage = &minWage; //the address of minWage is pointed to by p_minWage
	
	adjustMinWage(p_minWage); 
	/*why do I pass the value at p_minWage instead of the address? I.e. Why should it be *p_minWage instead of p_minWage?
	since the & (address operator) is in the function prototype, I'm lead to believe that I should be passing an address 
	and passing a pointer makes sense to me because it's value is an address
	*/
	cout << "The minimum wage has been adjusted to: $" << minWage << endl; 
	return 0;
}

void adjustMinWage(double &minWage){
	cout << "What would you like to change the new minimum wage to? $";
	cin >> minWage; 
}

I understand that the code in the second example fails to compile because the argument should be *p_minWage, but I don't understand why. Since the parameters of my prototype are (double &minWage), I'm lead to believe that I should be passing an address that contains a double. Pointers are assigned address that point to values, so passing the pointer or the address itself seems like the right idea to me. However, after trial and error, I should be using the deference operator * to pass the value being pointed to, but why is this?

In example 1 it was very clear what was happening. The prototype had a pointer, which to me makes it apparent that the argument being passed to the function had to have been an address. In example 2, a value is being passed to the function even though it's using the address of operator in the prototype. Why do we need to pass the value pointed to by the pointer instead of the pointer itself?

[Thomas1965]

In the second example you specify a reference to a double.
void adjustMinWage(double &minWage); //what does this function take?

To call it correctly adjustMinWage(minWage);

Have a look here to about the differences between pointer and references.
http://stackoverflow.com/questions/57483/what-are-the-differences-between-a-pointer-variable-and-a-reference-variable-in

[codekiddy]

I think you misunderstood what & means in function declaration:

void adjustMinWage(double &minWage)

it means you are passing a variable by reference and not by copy, the passed value will not be copied to the function body, instead the function will work with real value (the one located in main function).

you can still use pointer in main and pass by reference. but to do so you call the function by dereferencing a pointer:

adjustMinWage(*p_minWage);
here the function will work on real object, just like in first example you posted.