[cautious_int]

Because no truncation happens. In this case [] operator doesn't specify any type, only that the expression inside is an integer expression. While normally the type size_t is used for object and array sizes, [] takes any integer expression and the compiler won't complain.

[Dylan16807]

I'm talking about this line

  uint8_t *buffer = new (std::nothrow) uint8_t[size + chunk_size];

size + chunk_size is clearly unsafe to truncate to 32 bits, but it truncates anyway. When I say 'inside the new operator' I'm including the allocation function. Something truncates it. If it actually allocated 8GB, or failed to allocate 8GB, there would be no exploit.

[cautious_int]

I was talking about the same line.

Apparently new is a "special" operator, or there is a bug in the compiler. I also can't get a warning with g++.

The problem seems to be that, as I said, [] takes any integer expression, it is there where the value gets truncated when operator sizeof or new is applied on it since they either return or take a size_t value.