## Sunday, February 21, 2016

### A formal definition of mutation independence

Louis Dionne poses the problem of move independence in the context of C++, that is, under which conditions a sequence of operations
```f(std::move(x));
g(std::move(x));
```
is sound in the sense that the first does not interfere with the second. We give here a functional definition for this property that can be applied to the case Louis discusses.
Let X be some type and functions f: X T×X and g: X Q×X. The impurity of a non-functional construct in an imperative language such as C++ is captured in this functional setting by the fact that these functions return, besides the output value itself, a new, possibly changed, value of X. We denote by fT and fX the projection of f onto T and X, respectively, and similarly for g. We say that f does not affect g if
gQ(x) = gQ(fX(x)) ∀xX.
If we define the equivalence relationship ~g in X as
x ~g y iff gQ(x) = gQ(y),
then f does not affect g iff
fX(x) ~g xxX
or
fX([x]g) ⊆ [x]gxX,
where [x]g is the equivalence class of x under ~g.
We say that f and g are mutation-independent if f does not affect g and g does not affect f, that is,
fX([x]g) ⊆ [x]g and gX([x]f) ⊆ [x]fxX,
The following considers the case of f and g acting on separate components of a tuple: suppose that X = X1×X2 and f and g depend on and mutate X1 and X2 alone, respectively, or put more formally:
fT(x1,x2) = fT(x1,x'2),
fX2(x1,x2) = x2,
gQ(x1,x2) = gQ(x'1,x2),
gX1(x1,x2) = x1
for all x1, x'1X1, x2, x'2X2. Then  f and g are mutation-independent (proof trivial). Getting back to C++, given a tuple x, two operations of the form:
```f(std::get<i>(std::move(x)));
g(std::get<j>(std::move(x)));
```
are mutation-independent if i!=j; this can be extended to the case where f and g read from (but not write to) any component of x except the j-th and i-th, respectively.