IQOD #6

IQOD #6

Code:

void main()
{
      char *p;
      char buf[10] = {1,2,3,4,5,6,9,8};
      p = (buf+1)[5];
      printf("%d", p);
}

is there anything wrong with this code? If so, fix it and then whats the output? (remember, dont copy & paste in your compiler lol fix it 'on paper')

should move into the harder ones... these seem so trivial lol

Hmm lets see

what does it do?

Originally Posted by jimmydang09

what does it do?

It doesn't do much. It is a test problem. It's for fun v_v

Wow 4,5,and 6! I haven't had a chance to get on my computer all day D: Ummm random guesses for all three: 3, George Washington, Missouri

Originally Posted by B1ackAnge1

Code:

void main()
{
      char *p;
      char buf[10] = {1,2,3,4,5,6,9,8};
      p = (buf+1)[5];
      printf("%d", p);
}

void main()
{
int *p = new int;
char buf[10] = {1,2,3,4,5,6,9,8};
p = (int*)(buf+1)[5];
printf("%d", p);
delete p;
}

output is 9.

p=(buf+1)[5]; would make p point to 0x000009(which isnt what we want i guess)
int *p = new int creates a pointer to a newly allocated int which then gets assigned 9(which is what we want) and delete the allocated memory afterwards

The problem with:
char* p = (buf+1)[5]; is that (buf+1)[5]; returns a 'char' and not a pointer
so indeed the fix is to either make char*p a char instead, or cast the right hand operand.
The value printed is '9' since buf+1[5] would be the same as buf[6];

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