C++ Templates - The Complete Guide, 2nd Edition introduces the max template:
template<typename T>
T max (T a, T b)
{
// if b < a then yield a else yield b
return b < a ? a : b;
}
And it explains using “b < a ? a : b” instead of “a < b ? b : a”:
Note that the max() template according to [StepanovNotes] intentionally returns “b < a ? a : b” instead of “a < b ? b : a” to ensure that the function behaves correctly even if the two values are equivalent but not equal.
How to understand "even if the two values are equivalent but not equal."? “a < b ? b : a” seems have the same result for me.
a and b are equivalent, then !(a < b) && !(b < a) is true, so a < b and b < a are both false, so in b < a ? a : b, b is returned, which is not what you want... You want a < b ? b : a.
a = max(a, b); (repeatedly) you might not want to replace a unnecessarily.
a with a copy of a).
std::addressof is irrelevant. In fact, for the given T max(T a, T b) we already know addressof(a) != addressof(b).
std::max(a, b) is indeed specified to return a when the two are equivalent.
That's considered a mistake by Stepanov and others because it breaks the useful property that given a and b, you can always sort them with {min(a, b), max(a, b)}; for that, you'd want max(a, b) to return b when the arguments are equivalent.
This answer explains why the given code is wrong from a C++ standard point-of-view, but it is out of context.
See @T.C.'s answer for a contextual explanation.
The standard defines std::max(a, b) as follows [alg.min.max] (emphasis is mine):
template
Equivalent here means that !(a < b) && !(b < a) is true [alg.sorting#7].
In particular, if a and b are equivalent, both a < b and b < a are false, so the value on the right of : will be returned in the conditional operator, so a has to be on the right, so:
a < b ? b : a
...seems to be the correct answer. This is the version used by libstdc++ and libc++.
So the information in your quote seems wrong according to the current standard, but the context in which it is defined might be different.
X).
a<b and b<a can both be false because they're unordered (one or both NaN, so == is false too). That could be seen as a sort of equivalence. loosely related: x86's maxsd a, b instruction implements a = max(b,a) = b < a ? a : b. (What is the instruction that gives branchless FP min and max on x86?). The instruction keeps the source operand (the 2nd one) on unordered, so a loop over an array will give you NaN if there were any NaNs. But max_seen = max(max_seen, a[i]) will ignore NaNs.
The point is which one should be returned when they are equivalent; std::max has to return a (i.e. the first argument) for this case.
If they are equivalent, returns a.
So a < b ? b : a should be used; on the other hand, b < a ? a : b; will return b incorrectly.
(As @Holt said, the quote seems opposite.)
"the two values are equivalent but not equal" means they have the same value when being compared, but they migth be different objects at some other aspects.
e.g.
struct X { int a; int b; };
bool operator< (X lhs, X rhs) { return lhs.a < rhs.a; }
X x1 {0, 1};
X x2 {0, 2};
auto x3 = std::max(x1, x2); // it's guaranteed that an X which cantains {0, 1} is returned
std::max(a, b) has to return a, if a and b are equivalent?
a and b are equivalent, then !(a < b) && !(b < a) is true, so a < b and b < a are false, so... ?
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{min(a, b), max(b, a)}?max(a,b)will return a if-and-only-ifmin(a,b)returns b, and vice versa so that they are the reverse of each other and the (unordered) set{min(a,b), max(a,b)}is always equal to{a,b}.minandmaxto anything but the timestamp (the sort key) in that scenario makes no sense. The events (the objects) themselves shouldn't even be comparable if equality doesn't imply interchangeability. The only way{min(a, b), max(a, b)}makes any sense as a sort mechanism is if the objects are interchangeable.