为什么为传递给函数的指针获取不同的内存地址？

Question

Basically my question is, when I run these two segments of code, I get different memory addresses. The first segment of code gives a certain memory address for rValue, and the second gives a different memory address for rValue, just by adding a & operator. Why does this happen?

#include <iostream>
using namespace std;


int pMem(int *rValue){

cout << "Value of rValue is " << *rValue << endl;;
cout << "Address of rValue is " << rValue << endl;
*rValue = 15;
cout << "Value of rValue now is " << *rValue << endl;
cout << "Address of rValue is " << rValue << endl;

return *rValue;

}

int main() {


int value = 8;
int *pValue = &value;

pMem(&value);



cout << "Value = " << value << endl;
cout << "Address of Value: " << pValue << endl;
cout << "Value at memory address " << pValue << ": " << *pValue << endl;


return 0;
}

2nd block of code, this time with the &rValue... I get a different memory address than the first block of code.

#include <iostream>
using namespace std;


int pMem(int *rValue){

cout << "Value of rValue is " << *rValue << endl;;
cout << "Address of rValue is " << &rValue << endl;
*rValue = 15;
cout << "Value of rValue now is " << *rValue << endl;
cout << "Address of rValue is " << &rValue << endl;

return *rValue;

}

int main() {


int value = 8;
int *pValue = &value;

pMem(&value);



cout << "Value = " << value << endl;
cout << "Address of Value: " << pValue << endl;
cout << "Value at memory address " << pValue << ": " << *pValue << endl;


return 0;

}

Answer 1

Even pointers themselves take up memory and have an address associated with them.

So &rValue is the address of the the pointer rValue, and, unless it's a pointer to a function, that address will be different.

Answer 2

The reason for this behavior is that pointers are passed by value. In other words, when your function receives a pointer-typed parameter rValue like this

int pMem(int *rValue)

C++ allocates space for a brand-new variable of type int* , complete with its own address in memory. This variable gets initialized from a pointer expression that you pass to pMem . Other than that, it is a separate variable that behaves like a local variable to pMem . In particular, any re-assignments of the parameter itself have no effect on the caller.

It is the address of that parameter variable that gets printed when you do this:

cout << &pValue << endl; // Prints a new address

If you would like to see the address that you passed to the function, print the pointer, not the address of it:

cout << pValue << endl;  // Prints the address that you passed