How to use if-constexpr without repeating code?

Question

Currently I am doing:

if constexpr(constexpr_bool_var1) {
    auto arg1 = costly_arg1(); auto arg2 = costly_arg2();
    if (costly_runtime_function(arg1, arg2)) {
        // do X, possibly more constexpr conditions
        // do Y
        // ...
    }
} else {
    // do X, possibly more constexpr conditions
    // do Y
    // ...
}

One possible way is to convert the do X/Y etc. to one function doXY() and call it in both places, however it seems very unwieldy, as I have to write a function that solely exists for convenience of meta programming.

What I want is something like:

if not constexpr(constexpr_bool_var1 && some_magic(costly_runtime_function(arg1, arg2)) {
  // do X, do Y
}

Another way is:

auto arg1 = costly_arg1(); // Unneeded extra work out not within constexpr
auto arg2 = costly_arg2();
if (constexpr_bool_var1 && costly_runtime_function(arg1, arg2)) {
} else {
    // do X, possibly more constexpr conditions
    // do Y
    // ...
}

However here arg1 and arg2 are being declared outside the if condition so they will be needlessly instantiated.

What about a lambda outside the if? auto doXY = [] { doX(); doY(); }; and then call it in both places. — Praetorian
– Praetorian, Commented Oct 13, 2017 at 3:59
Since if constexpr requires the test to be done at compile time, and the result of costly_runtime_function() will be unknown until run time, it can't be used in if constexpr. — Peter
– Peter, Commented Oct 13, 2017 at 4:06
If I understand correctly, you want to do X and Y no matter what if constexpr_bool_var1 is false, and maybe if it's true but you won't know until runtime? In that case, can't you just do bool do_x_y = true; if constexpr (constexpr_bool_var1) { do_x_y = costly_runtime_function(/* ... */); } if (do_x_y) { /* ... */ }? — Daniel H
– Daniel H, Commented Oct 13, 2017 at 5:48
@DanielH - I think you should present your comment as an answer (a good answer, IMHO). — max66
– max66, Commented Oct 13, 2017 at 9:34
don't forget that if a value is evaluated at compile-time, condition like if (constexpr_val && runtime_val) is optimized into if (runtime_val) anyway. if constexpr is mostly meant to be used when suppressed branch cannot be compiled. — Andrei R.
– Andrei R., Commented Oct 13, 2017 at 9:36

Daniel H · Accepted Answer · 2017-10-13 14:21:39Z

3

I’m not sure I understand your question correctly; your original code and your second alternative don’t express quite the same function (the meaning of costly_runtime_function is reversed from “do X and Y” to “don’t do X and Y”), and in your first propoed alternative I don’t understand what your proposed syntax or what some_magic is. I’m answering your question with the semantics of your original code sample.

The best way to handle this is probably with a flag for whether or not to do X and Y:

bool do_x_y = true;
if constexpr(constexpr_bool_var1) {
    // Maybe we don't actually want X and Y
    auto arg1 = costly_arg1(); auto arg2 = costly_arg2();
    do_x_y = costly_runtime_function(arg1, arg2);
}
if (do_x_y)  {
    // do X, possibly more constexpr conditions
    // do Y
    // ...
}

Note that, as Andrei R. points out in the comments, the compiler can probably handle the optimization anyway. This makes it clearer to a human reader that you want this handled at compile time, though.

answered Oct 13, 2017 at 14:21

Daniel H

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1 Comment

themagicalyang Over a year ago

Thank you, I think this should work and is clear to the reader also.

Richard Hodges · Accepted Answer · 2017-10-13 05:43:22Z

0

Here's one way. Whether it's more expressive/maintainable I cannot say.

#include <cstdlib>
#include <utility>

/*
 * simulation
 */
void doX() {}
void doY() {}

int costly_arg1() { return 1; }
int costly_arg2() { return 2; }

bool costly_runtime_function(int, int) { return rand() < RAND_MAX / 2; }

constexpr bool constexpr_bool_var1 = true;

/*
 * A functor that maybe does something
 */
template<class F>
struct maybe_do
{
    constexpr maybe_do(F&& f) : f(std::move(f)) {}

    constexpr void operator()() const 
    {
        if (enabled)
            f();
    }

    constexpr void enable(bool e) {
        enabled = e;
    } 

    F f;
    bool enabled = true;
};

int main()
{
    auto thing = maybe_do{
        [] {
            doX();
            doY();
        }
    };

    if constexpr(constexpr_bool_var1) 
    {
        thing.enable(costly_runtime_function(costly_arg1(), 
                                             costly_arg2()));
    }
    thing();
}

answered Oct 13, 2017 at 5:43

Richard Hodges

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3 Comments

Daniel H Over a year ago

The OP didn’t want to make a doXY function because it seemed like too much extra code just for this one case, so you propose a lot more boilerplate instead? Either way, you have a bug; thing starts out enabled and you never disable it anywhere.

Richard Hodges Over a year ago

@DanielH thing is disabled if you call thing.enable(false);

Daniel H Over a year ago

Oh, I see. I still think that’s far too much boilerplate if even extracting X and Y into their own function is too much, but but it looks like it’ll work.

Guillaume Racicot · Accepted Answer · 2017-10-13 15:08:34Z

0

You could use a lambda for that:

auto stuff = [&] {
    // do X, possibly more constexpr conditions
    // do Y
    // ...
};

if constexpr(constexpr_bool_var1) {
    auto arg1 = costly_arg1(); auto arg2 = costly_arg2();
    if (costly_runtime_function(arg1, arg2)) {
        stuff();
    }
} else {
    stuff();
}

And if your lambda can receive auto values, you could also pass variable of different types from inside the if scope.

answered Oct 13, 2017 at 15:08

Guillaume Racicot

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