Is floating point addition commutative in C++?
For floating point values, is it guaranteed that a + b is the same as1b + a?
I believe this is guaranteed in IEEE754, however the C++ standard does not specify that IEEE754 must be used. The only relevant text seems to be from [expr.add]#3:
The result of the binary + operator is the sum of the operands.
The mathematical operation "sum" is commutative. However, the mathematical operation "sum" is also associative, whereas floating point addition is definitely not associative. So, it seems to me that we cannot conclude that the commutativity of "sum" in mathematics means that this quote specifies commutativity in C++.
Footnote 1:
"Same" as in bitwise identical, like memcmp rather than ==, to distinguish +0 from -0. IEEE754 treats +0.0 == -0.0 as true, but also has specific rules for signed zero. +0 + -0 and -0 + +0 both produce +0 in IEEE754, same for addition of opposite-sign values with equal magnitude. An == that followed IEEE semantics would hide non-commutativity of signed-zero if that was the criterion.
Also, a+b == b+a is false with IEEE754 math if either input is NaN.
memcmp will say whether two NaNs have the same bit-pattern (including payload), although we can consider NaN propagation rules separately from commutativity of valid math operations.
Solution 1:
It is not even required that a + b == a + b. One of the subexpressions may hold the result of the addition with more precision than the other one, for example when the use of multiple additions requires one of the subexpressions to be temporarily stored in memory, when the other subexpression can be kept in a register (with higher precision).
If a + b == a + b is not guaranteed, a + b == b + a cannot be guaranteed. If a + b does not have to return the same value each time, and the values are different, one of them necessarily will not be equal to one particular evaluation of b + a.
Solution 2:
No, the C++ language generally wouldn't make such a requirement of the hardware. Only the associativity of operators is defined.
All kinds of crazy things do happen in floating-point arithmetic. Perhaps, on some machine, adding zero to an denormal number produces zero. Conceivable that a machine could avoid updating memory in the case of adding a zero-valued register to a denormal in memory. Possible that a really dumb compiler would always put the LHS in memory and the RHS in a register.
Note, though, that a machine with non-commutative addition would need to specifically define how expressions map to instructions, if you're going to have any control over which operation you get. Does the left-hand side go into the first machine operand or the second?
Such an ABI specification, mentioning the construction of expressions and instructions in the same breath, would be quite pathological.