#### Evaluated Slots

The evaluated slot mechanism is a tool to fully evaluate a constant integral expression and avoid the lazy evaluation normally performed by the preprocessor.

#### Tutorial

In order to understand the use of such a mechanism, I will start with a simple file-iteration example. Consider the following scenario....

    for (int i = 0; i < 10; ++i) {
       for (int j = 0; j < i; ++j) {
          // ... use i and j
       }
    }

The above is a simple runtime model of the following multidimensional file-iteration....

    // file.hpp
    #if !BOOST_PP_IS_ITERATING
       #ifndef FILE_HPP_
       #define FILE_HPP_

       #include <boost/preprocessor/iteration/iterate.hpp>

       #define BOOST_PP_ITERATION_PARAMS_1 (3, (0, 9, "file.hpp"))
       #include BOOST_PP_ITERATE()

       #endif // FILE_HPP_
    #elif BOOST_PP_ITERATION_DEPTH() == 1
       #define I BOOST_PP_ITERATION()

       #define BOOST_PP_ITERATION_PARAMS_2 (3, (0, I, "file.hpp"))
       #include BOOST_PP_ITERATE()

       #undef I
    #elif BOOST_PP_ITERATION_DEPTH() == 2
       #define J BOOST_PP_ITERATION()

       // use I and J

       #undef J
    #endif

There is a problem with the code above. The writer expected I to refer the previous iteration frame. However, that is not the case. When the user refers to I, he is actually referring to BOOST_PP_ITERATION(), not the value of BOOST_PP_ITERATION() at the point of definition. Instead, it refers to exactly the same value to which J refers.
The problem is that the preprocessor always evaluates everything with lazy evaluation. To solve the problem, we need I to be evaluated here:

    // ...
    #elif BOOST_PP_ITERATION_DEPTH() == 1
       #define I BOOST_PP_ITERATION()
    // ...

Fortunately, the library offers a mechanism to do just that: evaluated slots. The following code uses this mechanism to "fix" the example above...

    // ...
    #elif BOOST_PP_ITERATION_DEPTH() == 1
       #define BOOST_PP_VALUE BOOST_PP_ITERATION()
       #include BOOST_PP_ASSIGN_SLOT(1)
       #define I BOOST_PP_SLOT(1)
    // ...

There are two steps to the assignment of an evaluated slot. First, the user must define the named external argument BOOST_PP_VALUE. This value must be an integral constant expression. Second, the user must include BOOST_PP_ASSIGN_SLOT(x), where x is the particular slot to be assigned to (1 to BOOST_PP_LIMIT_SLOT_COUNT). This will evaluate BOOST_PP_VALUE and assign the result to the slot at index x.
To retrieve a slot's value, the user must use BOOST_PP_SLOT(x).
In the case above, I is still lazily evaluated. However, it now evaluates to BOOST_PP_SLOT(1). This value will not change unless there is a subsequent call to BOOST_PP_ASSIGN_SLOT(1).

#### Advanced Techniques

The slot mechanism can also be used to perform calculations:

    #include <iostream>

    #include <boost/preprocessor/slot/slot.hpp>
    #include <boost/preprocessor/stringize.hpp>

    #define X() 4

    #define BOOST_PP_VALUE 1 + 2 + 3 + X()
    #include BOOST_PP_ASSIGN_SLOT(1)

    #undef X

    int main(void) {
       std::cout
          << BOOST_PP_STRINGIZE(BOOST_PP_SLOT(1))
          << &std::endl;
       return 0;
    }

In essence, anything that can be evaluated in an #if (or #elif) preprocessor directive is available except the defined operator.
It is even possible to use a particular slot itself while reassigning it:

    #define BOOST_PP_VALUE 20
    #include BOOST_PP_ASSIGN_SLOT(1)

    #define BOOST_PP_VALUE 2 * BOOST_PP_SLOT(1)
    #include BOOST_PP_ASSIGN_SLOT(1)

    BOOST_PP_SLOT(1) // 40

#### See Also

- BOOST_PP_ASSIGN_SLOT
- BOOST_PP_LIMIT_SLOT_COUNT
- BOOST_PP_SLOT
- BOOST_PP_VALUE

- Paul Mensonides