Is there a way to force a specific evaluation order of function arguments?

Question

I understand that when I call a function such as

a(b(),c());

then the behavior of this may be undefined in <= C++14, and unspecified in >= C++17, in the sense that it is up to the compiler to determine whether to evaluate b or c first.

I would like to know the best way to force an evaluation order. I will be compiling as C++14.

The thing that immediately comes to mind is something like this:

#include <iostream>

int count = 5;
auto increment(){
    return count++;
}

template <typename A, typename B>
auto diff(A && a, B && b){
   return a - b;
}

int main() {
    auto && a = increment();
    auto && b = increment();
    auto c = diff(a,b);
}

Am I in undefined behavior land? Or is this how one is "supposed" to force evaluation order?

Answer 1

The semi-colon that separates statements imposes a "happens before" relation. auto && a = increment() must be evaluated first. It is guaranteed. The returned temporary will be bound to the reference a (and its lifetime extended) before the second call to increment.

There is no UB. This is the way to force an evaluation order.

The only gotcha here is if increment returned a reference itself, then you'd need to worry about lifetime issues. But if there was no lifetime issues, say if it returned a reference to count, there still would not be UB from the imposed evaluation of a and then b.

Answer 2

Here's another way to force the evaluation order, using a std::initializer_list, which has a guaranteed left-to-right order of evaluation:

#include <numeric> // for accumulate
#include <initializer_list>

template <class T>
auto diff(std::initializer_list<T> args)
{
   return std::accumulate(args.begin(), args.end(), T(0), std::minus<>{});
}

const auto result = diff({increment(), increment()});

This restricts you to objects of the same type, and you need to type additional braces.