Why would they use this segment instead of this one?

Why use this segment

Code:

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#include <iostream>
using namespace std;

int main ()
{
  int numbers[5];
  int * p;
  p = numbers;  *p = 10;
  p++;  *p = 20;
  p = &numbers[2];  *p = 30;
  p = numbers + 3;  *p = 40;
  p = numbers;  *(p+4) = 50;
  for (int n=0; n<5; n++)
    cout << numbers[n] << ", ";
  return 0;
}

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instead of this segment?

Code:

---

#include <iostream>
using namespace std;

int main ()
{
  int a[5];
  int *b;
  b = a;  *b = 10;
  b = a;  *(b+1) = 20;
  b = a;  *(b+2) = 30;
  b = a;  *(b+3) = 40;
  b = a;  *(b+4) = 50;
  for (int n=0; n<6; n++)
    cout << a[n] << ", ";
  return 0;
}

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Does the first, seemingly more complicated block do something more than the second block? Even though both print the same thing?

NOTE in the first block, p is equivalent to b in the second block. and "numbers" is equivalent to a.

Obviously you would do it for educational purposes, to demonstrate various ways you can access an offset from a pointer.

By the way, the second version iterates one element too many on output.

In the real world, what you would do is

Code:

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int numbers[] = { 10, 20, 30, 40, 50  };

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and leave all the dangerous pointer stuff out.

Simply because they are slow, and take too many lines to do something pretty simple and easy:

Code:

---

int main()
{
  int a[5], *b = a;
  for(int i = 0; i < 5;)
      cout << (*(b+i) = (++i * 0xA)) << (char)0x20;
  return 0;
}

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Also, why not simply use a pointer from the start?

Code:

---

int main()
{
  int *a = new int[5];
  for(int i = 0; i < 5;)
      cout << (*(a+i) = ((i++ + 0x14 / 0x2 + i) * 0x5) - 0x28) << (char)0x20;
  return 0;
}

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you could also do it using inline asm, then it would get even more ureadable, if that is your goal.


As you can see, there is a lot of ways of making something more and more unreadable. Use your criativity.

Quote:

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Originally Posted by Dante Wingates

Simply because they are slow, and take too many lines to do something pretty simple and easy:

Code:

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  cout << (*(b+i) = (++i * 0xA)) << (char)0x20;

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This line causes undefined behaviour because some compilers will evaluate 'i' before the assigment and others will evaluate ++i after the assignment first