What happens when assigning a value to non-allocated memory?

Question

int main()
{
    char* p = new char('a');
    *reinterpret_cast<int*>(p) = 43523;

    return 0;
}

This code runs fine but how safe is it? It should have only allocated one byte of memory but it doesn't seem to be having any problem filling 4 bytes with data. What can happen with the other 3 bytes that are not allocated?

Answer 1

The code is not safe. Your program has undefined behaviour. Anything at all could happen.

Answer 2

The code causes heap buffer overflow. You're overriding memory which you don't own and can be in use in other parts of the program.

Answer 3

char* p = new char('a');

ASCII code of 'a' is 97, so this is equivalent to:

char* p = new char(97);

Single character (1 byte) space is allocated with *p='a' Now, you are trying to put more than 1-byte value, that's certainly risky, even if this works, I mean runs without any segmentation fault. You are overwriting some other parts of memory that you don't own or even if own, must be for some other purpose.