On Wednesday, 23 September 2015 at 03:39:02 UTC, Mike Parker
wrote:
I have a situation where I would like to demonstrate violating
the contract of immutable (as an example of what not to do),
but do so without using structs or classes, just basic types
and pointers. The following snippet works as I would expect:
```
immutable int i = 10;
immutable(int*) pi = &i;
int** ppi = cast(int**)π
writeln(*ppi);
int j = 9;
*ppi = &j;
writeln(*ppi);
```
Two different addresses are printed, so I've successfully
violated the contract of immutable and changed the value of pi,
an immutable pointer. However, this does not work as I expect.
```
immutable int x = 10;
int* px = cast(int*)&x;
*px = 9;
writeln(x);
```
It prints 10, where I expected 9. This is on Windows. I'm
curious if anyone knows why it happens.
violating immutable is undefined behaviour, so the compiler is
technically speaking free to assume it never happens. At the very
least, neither snippet's result is guaranteed to show a change or
not. At the most, literally anything can happen.
