[spc476]

What happens with this code?

    volatile int x;
    int          y;
    int          z;
    
    x = 10;
    x = 20;
    y = x;
    z = x;

Answer:

    the constant 10 is written to x
    the constant 20 is written to x
    the contents of x is read and written into y
    the contents of x is read and written into z

Now, what happens with this code?

    int x;
    int y;
    int z;
    
    x = 10;
    x = 20;
    y = x;
    z = x;

One answer is the same as the above. Another valid answer is:

    the constant 20 is written to x
    the constant 20 is written to y
    the constant 20 is written to z

Why? Because x is not used between the two assignments, so the first will never be seen. Also, x is not used between it's assignment and the assignment to y, so the compiler can do constant propagation.

All volatile does it tell the compiler "all writes must happen, and no caching of reads".

[tempguy9999]

Understood but we're talking about different things I think (though this is very much not my area).

You're saying volatile is acting as a kind of memory barrier instruction for the compiler - got it. But I'm saying I understand that at the CPU level, just considering x86 instructions, writes don't have to be forced to RAM, despite a common assumption that they are; they can remain in caches. See johntb86's reply confirming this.