Dynamic array confusion

Question

I was just brushing up my C concepts a bit where I got confused about some behavior. Consider the following code snippet:

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

int main (){


    int * arr;
    arr= malloc(3*sizeof(*arr));
    arr[0]=1;
    arr[1]=2;
    arr[2]=3;
    arr[3]=4;
    printf("value is %d \n", arr[3]);

return 0;


}

The problem is that the program functions correctly! As far as I understand I allocate memory for an array of 3 integers. So basically when I try to put a value in arr[3] there should be a segmentation fault as no memory has been assigned for it. But it works fine and prints the value 4. Either this is some weird behavior or I seriously need to revise basic C. Please if anyone can offer some explanation I would highly appreciate it. Thanks.

Answer 1

Technically, It is Undefined Behavior, Which means anything can happen not necessarily a segmentation fault.
Just your program is not a valid program and you should not write invalid programs and expect valid /invalid behaviors from them.

Answer 2

You could get a segmentation fault at any time, you got "lucky" this time. This is undefined behavior, so you might get a seg fault some other time.

C does not do any bounds checking, so while e.g., Java would have complained, C is quite happy to do whatever you ask it to do, even to the program's own detriment.

This is both one of its major strengths, and also weaknesses.