How is the three-way comparison operator different from subtraction?

问题

There's a new comparison operator <=> in C++20. However I think in most cases a simple subtraction works well:

int my_strcmp(const char *a, const char *b) {
    while (*a == *b && *a != 0 && *b != 0) {
        a++, b++;
    }
    // Version 1
    return *a - *b;
    // Version 2
    return *a <=> *b;
    // Version 3
    return ((*a > *b) - (*a < *b));
}

They have the same effect. I can't really understand the difference.

回答1:

The operator solves the problem with numeric overflow that you get with subtraction: if you subtract a large positive number from a negative that is close to INT_MIN, you get a number that cannot be represented as an int, thus causing undefined behavior.

Although version 3 is free from this problem, it utterly lacks readability: it would take some time to understand by someone who has never seen this trick before. <=> operator fixes the readability problem, too.

This is only one problem addressed by the new operator. Section 2.2.3 of Herb Sutter's Consistent comparison paper talks about the use of <=> with other data types of the language where subtraction may produce inconsistent results.

回答2:

Here are some cases that subtraction won't work for:

1. unsigned types.
2. Operands that cause integer overflow.
3. User-defined types that don't define operator - (perhaps because it's not meaningful - one may define an order without defining a notion of distance).

I suspect this list is non-exhaustive.

Of course, one can come up with workarounds for at least #1 and #2. But the intent of operator <=> is to encapsulate that ugliness.

回答3:

There are some meaningful answers here on the difference, but Herb Sutter in his paper specifically says:

<=> is for type implementers: User code (including generic code) outside the implementation of an operator<=> should almost never invoke an <=> directly (as already discovered as a good practice in other languages);

So even if there was no difference, the point of the operator is different: to aid class writers to generate comparison operators.

The core difference between the subtraction operator and the "spaceship" operator (according to Sutter's proposal) is that overloading operator- gives you a subtraction operator, whereas overloading operator<=>:

• gives you the 6 core comparison operators (even if you declare the operator as default: no code to write!);
• declares whether your class is comparable, is sortable, and whether the order is total or partial (strong/weak in Sutter's proposal);
• allows for heterogeneous comparisons: you can overload it to compare your class to any other type.

Other differences are in the return value: operator<=> would return an enum of a class, the class specifies whether the type is sortable and whether the sort is strong or weak. The return value would convert to -1, 0 or 1 (though Sutter leaves room for the return type to also indicate distance, as strcmp does). In any case, assuming the -1, 0, 1 return value, we'll finally get a true signum function in C++! (signum(x) == x<=>0)

来源：https://stackoverflow.com/questions/48042955/how-is-the-three-way-comparison-operator-different-from-subtraction