Help: double free corruption??

Code:

---

int main()
{
        int *a,i;
        a=(int *)malloc(10*sizeof(int));
        for(i=0;i<10;i++)
                *(a+i)=i*i;
        for(i=0;i<10;i++)
                printf("%d",*a++);
        free(a);
return 0;
}

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why this code is showing double free corruption?

It may not be a double free per se, but you are incrementing a. Rather, have another pointer to be a copy of a, and increment that. That way, free(a) will operate on the original location pointed to by a.

This smells like a test question.

What is a? What is its value prior to the final loop? What about after? How can that affect the free operation?

Thanks laser

I have used a+i in place of a++ it fine now.

But another doubt

Code:

---

main()
{
    double d;
    printf("%d", ((double *)0+2));
}

---

How this code is giving 16?

Because %d does NOT print doubles.

r2r is right, you're just trotting off quiz questions.

He's printing a pointer value there. 16 would be the expected offset from NULL.

Quote:

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Originally Posted by GOGO1104

Thanks laser

I have used a+i in place of a++ it fine now.

But another doubt

Code:

---

main()
{
    double d;
    printf("%d", ((double *)0+2));
}

---

How this code is giving 16?

---

It's not a doubt, it's a question.

I think whoever is teaching English in that part of the world needs to find another career. :)

Don't worry, I am not talking about you but of the person that taught you English.

%d does not print pointers either. That's undefined behavior.
Furthermore, main must have a return type of main and a return 0; at the end.