try to understand pointer

Hello! I was reading the tutorial about pointer on this website and I found a sentence that I don't quite understand after several times of reading. What does it mean? Can someone please help? Thank you! :)

here is the original test from the tutorial

*p++ = *q++;

Because ++ has a higher precedence than *, both p and q are increased, but because both increase operators (++) are used as postfix and not prefix, the value assigned to *p is *q before both p and q are increased. And then both are increased. It would be roughly equivalent to:

1
2
3

*p = *q;
++p;
++q;

Chervil (7320)

Pointers and arrays have a lot of similarities. If we think of the above example in terms of arrays instead of pointers, it might be a bit like this.

1
2
3

array1[j] = array2[k];
j = j + 1;
k = k + 1;

that is, an element is copied from one array to another, and then each subscript is incremented, ready to access the next element.

Smac89 (1727)

#include <iostream>
using std::cout;
using std::endl;
int main() {
    int a[] = {1, 2, 3, 4}, b[] = {5, 6, 7, 8};
    
    cout << "Before assignment\n";
    int *p = a, *q = b;
    cout << "Value of *p is " << *p << " and value of *q is " << *q << endl;
    
    cout << "After assignment -> *p++ = *q++\n";
    *p++ = *q++;
    cout << "Value of *p is " << *p << " and value of *q is " << *q << endl;
    
    cout << "Moving the pointers back to original position... -> *--p, *--q\n";
    cout << "Value of *p is " << *--p << " and value of *q is " << *--q << endl;
    return 0;
}

See c++ operator precedence for more information
http://en.cppreference.com/w/cpp/language/operator_precedence

Last edited on

long double main (999)

Let's put some context to that (excuse my variable names; I was typing this up quickly):

#include <iostream>
using std::cout;

int main()
{
    int a[] = {1,2,3,4,5,6,7};
    int* p = a, *q = a+3;
    cout << "The array before:\n";
    for (int i : a)
        cout << i << ' ';
    cout << "\nPosition of p before: " << (p-a);
    cout << "\nPosition of q before: " << (q-a);
    *p++ = *q++;
    cout << "\n\nArray after:\n";
    for (int i : a)
        cout << i << ' ';
    cout << "\nPosition of p after: " << (p-a);
    cout << "\nPosition of q after: " << (q-a);
    
}

The array before:
1 2 3 4 5 6 7
Position of p before: 0
Position of q before: 3

Array after:
4 2 3 4 5 6 7
Position of p after: 1
Position of q after: 4

Now, *p++ = *q++; is the same as *(p++) = *(q++), since ++ has a higher precedence than * (which means the ++ happens before the *).
Before this, p is pointing at the first element of a and q is pointing at the 4^th element of a.
Now, q++ means "increment the position of q and give me back the old value", so *q++ increments q and gives you the value in the array at the old position.
Basically, it's the same as doing *q followed by q++; later on.
So *q++ returns the element in the fourth position of a (which is 4) and then increments q (so now q is pointing at the fifth element of a).

The same thing happens with *p++ -- it increments p (to make it point at the 2nd position of a now) and then gives you the array element at the old position of p (which is 1). In this case, it's setting that element equal to the right-hand side of that expression, which we just figured out above to be 4.

So basically, all you really have to get out of it is this:
1) *p++ = *q++ is the same as

1
2
3

*p = *q;
++p;
++q;

2) Use parenthesis to avoid confusion. In this case, it would have been much clearer if it had been written as *(p++) = *(q++) rather than *p++ = *q++.

2) Sometimes clarity is better than brevity. Write code that you can actually understand (and will be able to understand when you look back at it 6 months afterwards).
(Hopefully, that goes without saying anyways, but....)

Topic archived. No new replies allowed.

C++

Forum

try to understand pointer