Problem with Dynamic Array

Hi All.. I cannot find the mistake in this code.

Code:

---

#include <stdio.h>
#include <stdlib.h>

void create_vect (int** v) {
  *v = (int *) malloc(2 * sizeof(int));
  *v[0] = *v[1] = 85;
}


int main()
{
    int* v;
        create_vect(&v);
        printf("%d %d\n", v[0], v[1]); //it prints "85 0". Why? It should be "85 85"

        return 0;
}

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PS: I don't want to change my function to "int create_vect (void)". I'd like to know why it doesn't work in THIS way

Thank you in advance !!!!

Because of operator precedence, you need to change this:

Code:

---

*v[0] = *v[1] = 85;

---

to this:

Code:

---

(*v)[0] = (*v)[1] = 85;

---

[] has higher precedence than *, so an expression like *v[0] evaluates to *(v[0]). But if you think about it, you need (*v)[0]. To get that, you add the parentheses.

Here's an operator precedence table in case you're interested (it's kind of technical): http://cppreference.com/operator_precedence.html

Oh my god !!!

Really Fast Reply :D

I'm so stupid.. I didn't know about that precedence :(

Anyway, Thank you very much :) Now it woks !

I've del.icio.used your website :D

Yes, you tend to get very fast replies on this board, depending on when you post the question. Sometimes you could get three people answering your question within as many minutes.

I don't update mny website very often . . . for a while there I uploaded codeform version 1.2.0 to it at least twice a week, but now it doesn't get changed very much. ;)

a fast and easy c reference guide is what i needed :)

nice work :D

Don't cast the return value of malloc(), there is no need.