CODEWITHSUNDEEP
c++
Radhi
Radhi

Reputation: 9

Passing pointer to calling function but called function parameter as reference

I have tried this piece of code mentioned below, I'm not understanding why *p need to be pass to doOperation() function. Why can't we pass p? What is the difference between the two?

doOperation(*p); // need explanation why derefencing
doOperation(p); // Gives compilation error

int main()
{
    int *p = getpointer();
    std::cout << "p:" << *p << std::endl;
    doOperation(*p); // Why this has to be pass as a pointer when the function parameter as reference
    return 0;
}

void doOperation(int &ptr) 
{
   //std::cout << "ptr:" << ptr << std::endl; 
}

int *getpointer()
{
    int *ptr = new int[10];
    int i;
    for (i=0; i <= 10; i++)
    {
        *(ptr+i) = i;
    }

    return ptr;
}

Upvotes: 1

Views: 40

Answers (1)

Osama Ahmad
Osama Ahmad

Reputation: 2096

You've declared p as an integer pointer. The function doOperation takes an int reference as a parameter. doOperation(*p) means that you're dereferencing the pointer (which points to the first element in the array) and passing it to the function. Also as @dxiv have pointed out, in the function getpointer, the loop initializes 11 elements instead of 10. you can solve this by just changing <= to <.

If you want to pass the pointer by reference instead, the function doOperation can look like this:

void doOperation(int *&ptr)
{
    std::cout << "ptr:" << ptr << std::endl;
}

Then you can just pass the pointer as an argument like this:

doOperation(p);

Output:

p:0
ptr:0x2b71550

The code after the changes should look like this:

#include <iostream>

void doOperation(int *&ptr)
{
    std::cout << "ptr:" << ptr << std::endl;
}

int *getpointer()
{
    int *ptr = new int[10];
    int i;
    for (i=0; i < 10; i++)
    {
        *(ptr+i) = i;
    }

    return ptr;
}

int main()
{
    int *p = getpointer();
    std::cout << "p:" << *p << std::endl;
    doOperation(p);
    return 0;
}

Upvotes: 1

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