2 ms·

To elaborate, we treat pointers as more than just integers because it gives optimizers the latitude to reorder and eliminate pointer operations. In the example

by kibwen 2y ago

To elaborate, we treat pointers as more than just integers because it gives optimizers the latitude to reorder and eliminate pointer operations. In the example above we cannot do this, because we cannot prove at compile time that x doesn't live at the address returned by oracle. For some high-quality further discussion, see Ralf Jung's series of blog posts starting with https://www.ralfj.de/blog/2018/07/24/pointers-and-bytes.html https://www.ralfj.de/blog/2018/07/24/pointers-and-bytes.html

shultays 2y ago
However, given how low-level a language C++ is, we can actually break this assumption by setting i to y-x. Since &x[i] is the same as x+i, this means we are actually writing 23 to &y[0]. But that is undefined, you can't do x + (y - x) ie a pointer arithmetic that ends outside of bounds of an array. Since it is undefined, shouldn't C++ assume that changing x[..] can't change y[0] edit: welp, if I read a few more lines into article I would see that it also tells it is undefined
gpderetta 2y ago
to be clear, in my example the result of oracle() cannot possibly alias with 'x' in C or C++ (and in fact gcc will optimize accordingly). In a different language where addresses are mere integers, things would be more complicated.
- codedokode 2y ago
  The result of oracle can point to anything if you write it as return (int *)rand(); Note that rand() returns 32-bit value so you have to call it twice and merge the results to obtain a 64-bit pointer.
  - gpderetta 2y ago
    The numerical value returned by oracle might physically match the address of the stack slot for 'x', assuming that it exists, but it doesn't mean that, from a language point of view, it is a valid pointer. If forging pointers had defined behaviour, it would be impossible to use the language sanely or perform any kind of optimization.

2 ms·

To elaborate, we treat pointers as more than just integers because it gives optimizers the latitude to reorder and eliminate pointer operations. In the example

by kibwen 2y ago

shultays 2y ago
However, given how low-level a language C++ is, we can actually break this assumption by setting i to y-x. Since &x[i] is the same as x+i, this means we are actually writing 23 to &y[0]. But that is undefined, you can't do x + (y - x) ie a pointer arithmetic that ends outside of bounds of an array. Since it is undefined, shouldn't C++ assume that changing x[..] can't change y[0] edit: welp, if I read a few more lines into article I would see that it also tells it is undefined
gpderetta 2y ago
to be clear, in my example the result of oracle() cannot possibly alias with 'x' in C or C++ (and in fact gcc will optimize accordingly). In a different language where addresses are mere integers, things would be more complicated.
- codedokode 2y ago
  The result of oracle can point to anything if you write it as return (int *)rand(); Note that rand() returns 32-bit value so you have to call it twice and merge the results to obtain a 64-bit pointer.
  - gpderetta 2y ago
    The numerical value returned by oracle might physically match the address of the stack slot for 'x', assuming that it exists, but it doesn't mean that, from a language point of view, it is a valid pointer. If forging pointers had defined behaviour, it would be impossible to use the language sanely or perform any kind of optimization.