Is a pointer passed by value?

Question

https://www.sololearn.com/discuss/315095/?ref=app

Similar to this, I have tried creating various functions with the returning pointer as an argument and as a return value...

In all of the return cases, the new pointer got updated with the values...

But in most of the argument cases, the passed pointer remained blank, and didn't have any update at all...

So how to solve this? Help.

//This makes me conclude that pointers are passed by value, but that is impossible!

//Pointers are memory representations!

Answer 1

Consider:

#include <iostream>
using namespace std;

void ptr_alt (int *p)
{
p = new int;
}

int main()
{
int *p = NULL;

ptr_alt(p);

cout << p;

return 0;
}

// The result of pointer variable address will always be NULL.

Consider:

#include <iostream>
using namespace std;

void ptr_alt (int *&p)
{
p = new int;
}

int main()
{
int *p = NULL;

ptr_alt(p);

cout << p;

return 0;
}

// Here, we pass the pointer variable by reference. Result of p will always contain dynamically allocated address.

// Conclusion: Pointer variables are passed by value unless stated otherwise.

Answer 2

@rain_drop
BTW...

int main()
{
int no[100],esum=0,osum=0,n,a=0,b=0;
cout<<"How many Numbers? - ";cin>>n;
for(int i=0;i<n;i++)
{
cout<<"Enter Number "<<i+1<<" - ";
cin>>no[i];
if(no[i]%2==0)
{
e[a]=no[i];
esum+=no[i];
a++;
}
else
{
o[b]=no[i];
osum+=no[i];
b++;
}
}
cout<<"Even Numbers - ";
for(int j=0;j<a;j++)
{
cout<<e[j]<<",";
}
cout<<"\nSum - "<<esum<<endl;
cout<<"Odd Numbers - ";
for(int k=0;k<b;k++)
{
cout<<o[k]<<",";
}
cout<<"\nSum - "<<osum<<endl;
}

Answer 3

@rain_drop
How is it related to my question?

Answer 4

So, a pointer is passed by value after all...