malloc 接触未分配的内存

Question

I'm wondering why my allocation is working. I created ptr which is a pointer and ptr2 which a pointer on the same location than ptr .

Here's my code:

int* ptr = malloc(sizeof(int));
int* ptr2 = ptr;
*ptr2 = 1;
ptr2 = ptr+1;
*ptr2 = 2;

At the end I've an array of two integer like that :

ptr[0] = 1
ptr[1] = 2

My question is : why I can affect the value 2 to ptr2 without error or warnings ? I only malloc() -ed for one integer place in my array, isn't it?

Answer 1

The problem here is, by saying

  ptr2 = ptr+1;

you're pointing to out of bound memory. Next upon dererencing, you invoke undefined behavior . Now, the outcome of UB cannot be justified, in any way.

There is nothing inherent in C that stops you from accessing out of bound (or invalid) memory, but the outcome of doing so is UB.

That said, just a suggestion, you should always check the returned value of malloc() before using that pointer to avoid UB by dereferencing NULL pointer, in case of malloc() failure.