Using *(pointer+i) versus pointer[i]

Question

Can somebody please explain this code

// Option 1
int **p = new Point*[2];
p[1] = new Point;
p[1]->x = p[1]->x = 1;

// Option 2
int **p = new Point*[2];
*(p+1) = new Point;
(*(p+1))->x = (*(p+1))->x = 1;

Isn't both options the same? Why when I create the variable using option 2, I cant write its value with option 1 (I got random numbers (address numbers?))? Is there any difference?

Yes, there's a difference; the second example never initialises y. — Mr Lister
– Mr Lister, Commented Apr 24, 2012 at 18:45
Well, for one thing you don't assign y to anything in Option 2 -- you do ->x twice. — Nathan Monteleone
– Nathan Monteleone, Commented Apr 24, 2012 at 18:46
Oh, I see you edited your example now so that both options have the same error. — Mr Lister
– Mr Lister, Commented Apr 24, 2012 at 19:20
The pseudo-code you have provided doesn't demonstrate the problem. Please create the smallest actual program that does demonstrate the problem and paste that program, in its entirety, into your question. See sscce.org. (Hint: you should be able to demonstrate the problem in 10 lines or so.) — Robᵩ
– Robᵩ, Commented Apr 24, 2012 at 19:22
@Robᵩ the OP "demonstrates" the problem in the comment to CodeChordsman's answer below, by using some UB in printf. — Mr Lister
– Mr Lister, Commented Apr 24, 2012 at 19:25

Israel Unterman · Accepted Answer · 2012-04-24 18:46:11Z

1

You have a typo: (*(p+1))->x = (*(p+1))->x = 1; - should be y in the second term

answered Apr 24, 2012 at 18:46

Israel Unterman

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2 Comments

Buksy Over a year ago

sorry, there should by "x" as you have written. Im writing a project and I got different outputs on printf("%f != %f", p[1], *(p+1));

Mr Lister Over a year ago

@Buksy Why are you using %f there? %f expects a double, not a pointer to a struct. Of course it prints two different values!

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