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Quantity Arithmetics

quantity is a numeric wrapper

If we think about it, the quantity class template is just a "smart" numeric wrapper. It exposes properly constrained set of arithmetic operations on one or two operands.

Important: quantity propagates the underlying interface

Every single arithmetic operator is exposed by the quantity class template only if the underlying representation type provides it as well, and when its implementation has proper semantics (e.g., returns a reasonable type).

For example, in the following code, -a will compile only if MyInt exposes such an operation as well:

quantity a = MyInt{42} * m;
quantity b = -a;

Assuming that:

here is the list of all the supported operators:

  • unary:
    • +q
    • -q
    • ++q
    • q++
    • --q
    • q--
  • compound assignment:
    • q += qi
    • q -= qi
    • q %= qi
    • q *= number
    • q *= q1
    • q /= number
    • q /= q1
  • binary:
    • q + qk
    • q - qk
    • q % qk
    • q * qq
    • q * number
    • number * q
    • q / qq
    • q / number
    • number / q
  • ordering and comparison:
    • q == qk
    • q <=> qk

As we can see, there are plenty of operations one can do on a value of a quantity type. As most of them are obvious, in the following chapters, we will discuss only the most important or non-trivial aspects of quantity arithmetics.

Addition and subtraction

Quantities can easily be added or subtracted from each other:

static_assert(1 * m + 1 * m == 2 * m);
static_assert(2 * m - 1 * m == 1 * m);
static_assert(isq::height(1 * m) + isq::height(1 * m) == isq::height(2 * m));
static_assert(isq::height(2 * m) - isq::height(1 * m) == isq::height(1 * m));

The above uses the same types for LHS, RHS, and the result, but in general, we can add, subtract, or compare the values of any quantity type as long as both quantities are of the same kind. The result of such an operation will be the common type of the arguments:

static_assert(1 * km + 1.5 * m == 1001.5 * m);
static_assert(isq::height(1 * m) + isq::width(1 * m) == isq::length(2 * m));
static_assert(isq::height(2 * m) - isq::distance(0.5 * m) == 1.5 * m);
static_assert(isq::radius(1 * m) - 0.5 * m == isq::radius(0.5 * m));

Note

Please note that for the compound assignment operators, we always need to end up with the left-hand-side argument type:

static_assert((1 * m += 1 * km) == 1001 * m);
static_assert((isq::length(1 * m) += isq::height(1 * m)) == isq::length(1 * m));
static_assert((isq::height(1.5 * m) -= 1 * m) == isq::height(0.5 * m));

If we will break typical library's convertibility rules, the following code will not compile:

quantity q1 = 1 * m -= 0.5 * m;                         // Compile-time error(1)
quantity q2 = 1 * km += 1 * m;                          // Compile-time error(2)
quantity q3 = isq::height(1 * m) += isq::length(1 * m); // Compile-time error(3)
  1. Conversion of the floating-point to integral representation type is considered narrowing.
  2. Conversion of quantity with integral representation type from a unit of a higher resolution to the one with a lower resolution is considered narrowing.
  3. Conversion from a more generic quantity type to a more specific one is considered unsafe.

Multiplication and division

Multiplying or dividing a quantity by a number does not change its quantity type or unit. However, its representation type may change. For example:

static_assert(isq::height(3 * m) * 0.5 == isq::height(1.5 * m));

Note

Unless we use a compound assignment operator, in which case we always have to result with the type of the left-hand-side argument. This, together with the fact that this library tries to prevent truncation of a quantity value means, that the following does not compile:

quantity q = isq::height(3 * m) *= 0.5; // Compile-time error

However, suppose we multiply or divide quantities of the same or different types or we divide a raw number by a quantity. In that case, we most probably will end up in a quantity of yet another type:

static_assert(120 * km / (2 * h) == 60 * km / h);
static_assert(isq::width(2 * m) * isq::length(2 * m) == isq::area(4 * m2));
static_assert(50 / isq::time(1 * s) == isq::frequency(50 * Hz));

Note

An exception from the above rule happens when one of the arguments is a dimensionless quantity. If we multiply or divide by such a quantity, the quantity type will not change. If such a quantity has a unit one, also the unit of a quantity will not change:

static_assert(120 * m / (2 * one) == 60 * m);

An interesting special case happens when we divide the same quantity kinds or multiply a quantity by its inverted type. In such a case, we end up with a dimensionless quantity.

static_assert(isq::height(4 * m) / isq::width(2 * m) == 2 * one); // (1)!
static_assert(5 * h / (120 * min) == 0 * one);  // (2)!
static_assert(5. * h / (120 * min) == 2.5 * one);
  1. The resulting quantity type of the LHS is isq::height / isq::width, which is a quantity of the dimensionless kind.
  2. The resulting quantity of the LHS is 0 * dimensionless[h / min]. To be consistent with the division of different quantity types, we do not convert quantity values to a common unit before the division.

Important: Beware of integral division

The physical units library can't do any runtime branching logic for the division operator. All logic must be done at compile-time when the actual values are unknown, and the quantity types can't change at runtime.

If we expect 120 * km / (2 * h) to return 60 km / h, we have to agree with the fact that 5 * km / (24 * h) returns 0 km/h. We can't do a range check at runtime to dynamically adjust scales and types based on the values of provided function arguments.

This is why we often prefer floating-point representation types when dealing with units. Some popular physical units libraries even forbid integer division at all.

Modulo

Now that we know how addition, subtraction, multiplication, and division work, it is time to discuss modulo. What would we expect to be returned from the following quantity equation?

auto q = 5 * h % (120 * min);

Most of us would probably expect to see 1 h or 60 min as a result. And this is where the problems start.

C++ language defines its / and % operators with the quotient-remainder theorem:

q = a / b;
r = a % b;
q * b + r == a;

The important property of the modulo operation is that it only works for integral representation types (it is undefined what modulo for floating-point types means). However, as we saw in the previous chapter, integral types are tricky because they often truncate the value.

From the quotient-remainder theorem, the result of modulo operation is r = a - q * b. Let's see what we get from such a quantity equation on integral representation types:

const quantity a = 5 * h;
const quantity b = 120 * min;
const quantity q = a / b;
const quantity r = a - q * b;

std::cout << "reminder: " << r << "\n";

The above code outputs:

reminder: 5 h

And now, a tough question needs an answer. Do we really want modulo operation on physical units to be consistent with the quotient-remainder theorem and return 5 h for 5 * h % (120 * min)?

This is exactly why we decided not to follow this hugely surprising path in the mp-units library. The selected approach was also consistent with the feedback from the C++ experts. For example, this is what Richard Smith said about this issue:

Richard Smith

I think the quotient-remainder property is a less important motivation here than other factors -- the constraints on % and / are quite different, so they lack the inherent connection they have for integers. In particular, I would expect that A / B works for all quantities A and B, whereas A % B is only meaningful when A and B have the same dimension. It seems like a nice-to-have for the property to apply in the case where both / and % are defined, but internal consistency of / across all cases seems much more important to me.

I would expect 61 min % 1 h to be 1 min, and 1 h % 59 min to also be 1 min, so my intuition tells me that the result type of A % B, where A and B have the same dimension, should have the smaller unit of A and B (and if the smaller one doesn't divide the larger one, we should either use the gcd / std::common_type of the units of A and B or perhaps just produce an error). I think any other behavior for % is hard to defend.

On the other hand, for division it seems to me that the choice of unit should probably not affect the result, and so if we want that 5 mm / 120 min = 0 mm/min, then 5 h / 120 min == 0 hc (where hc is a dimensionless "hexaconta", or 60x, unit). I don't like the idea of taking SI base units into account; that seems arbitrary and like it would do the wrong thing as often as it does the right thing, especially when the units have a multiplier that is very large or small. We could special-case the situation of a dimensionless quantity, but that could lead to problematic overflow pretty easily: a calculation such as 10 s * 5 GHz * 2 uW would overflow an int if it produces a dimensionless quantity for 10 s * 5 GHz, but it could equally produce 50 G * 2 uW = 100 kW without any overflow, and presumably would if the terms were merely reordered.

If people want to use integer-valued quantities, I think it's fundamental that you need to know what the units of the result of an operation will be, and take that into account in how you express computations; the simplest rule for heterogeneous operators like * or / seems to be that the units of the result are determined by applying the operator to the units of the operands -- and for homogeneous operators like + or %, it seems like the only reasonable option is that you get the std::common_type of the units of the operands.

To summarize, the modulo operation on physical units has more in common with addition and division operators than with the quotient-remainder theorem. To avoid surprising results, the operation uses a common unit to do the calculation and provide its result:

static_assert(5 * h / (120 * min) == 0 * one);
static_assert(5 * h % (120 * min) == 60 * min);
static_assert(61 * min % (1 * h) == 1 * min);
static_assert(1 * h % (59 * min) == 1 * min);

Comparison against zero

In our code, we often want to compare the value of a quantity against zero. For example, we do it every time we want to ensure that we deal with a non-zero or positive value.

We could implement such checks in the following way:

if (q1 / q2 != 0 * m / s)
  // ...

The above would work (assuming we are dealing with the quantity of speed) but could be suboptimal if the result of q1 / q2 is not expressed in m / s. To eliminate the need for conversion, we need to write:

if (auto q = q1 / q2; q != q.zero())
  // ...

but that is a bit inconvenient, and inexperienced users could be unaware of this technique and its reasons.

For the above reasons, the library provides dedicated interfaces to compare against zero that follow the naming convention of named comparison functions in the C++ Standard Library. The mp-units/compare.h header file exposes the following functions:

  • is_eq_zero
  • is_neq_zero
  • is_lt_zero
  • is_gt_zero
  • is_lteq_zero
  • is_gteq_zero

Thanks to them, to save typing and not pay for unneeded conversions, our check could be implemented as follows:

if (is_neq_zero(q1 / q2))
  // ...

Tip

Those functions will work with any type T that exposes zero() member function returning something comparable to T. Thanks to that, we can use them not only with quantities but also with std::chrono::duration or any other type that exposes such an interface.

Other maths

This chapter scopes only on the quantity type's operators. However, there are many named math functions taking quantities as arguments. Those can be found in the mp-units/math.h header file. Among others, we can find there the following:

  • pow(), sqrt(), cbrt(),
  • exp(),
  • abs(),
  • epsilon(),
  • fma(), fmod(), remainder(),
  • isfinite(), isinf(), isnan(),
  • floor(), ceil(), round(),
  • inverse(),
  • hypot(),
  • sin(), cos(), tan(),
  • asin(), acos(), atan(), atan2().

In the library, we can also find mp-units/random.h header file with all the pseudo-random number generators working on quantity types.