A min= idiom in C++?

Question

We use

x += y

instead of

x = x + y

And similarly for *,/,- and other operators. Well, what about

Answer 1

NO. There is no such thing, you'll have to do with std::min(x,y);

Answer 2

You can't extend the language in this way. The closest you can come is something like:

template <typename T, typename U>
T&
mineq( T& lhs, U rhs )
{
    if ( rhs < lhs ) {
        lhs = rhs;
    }
    return lhs;
}

This would allow writing:

mineq( x, y );

I question whether it's worth the bother, however.

Answer 3

C++ has a limited set of operators and keywords.

What you are trying to do is outside the C++ specification and is not possible.

You can do the comparison and assignment with this one-liner if you want:

x = (x < y) ? x : y;

Answer 4

It's certainly not idiomatic, but you might be able to use something called named operators (see these Q&As here and here, developed by @Yakk and @KonradRudolph), and write

x <min>= y;

which is made possible by overloading operator< and operator>, combined with a clever wrapped named_operator. The full code is given by the link above, but uses code like

template <typename T1, typename T2, typename F>
inline auto operator >(named_operator_lhs<T1, F> const& lhs, T2 const& rhs)
    -> decltype(lhs.f(std::declval<T1>(), std::declval<T2>()))
{
    return lhs.f(lhs.value, rhs);
}

Using std::min as template argument for the template parameter F, would update the lhs of the expression with the min of the lhs and rhs.

Answer 5

You cannot write this kind of sentences, they are reserved for the built-in syntaxt

Answer 6

No, it is not possible to create new custom operators.

You have a few available solutions though:

llama_min_age = std::min(x, y);
llama_min_age = (x < y ? x : y);

Or even a macro if you want to:

#define MIN(x, y) ((x) < (y) ? (x) : (y))

About the macro: it can lead to vicious bug, so I would prefer to use one of the first two solutions.