是否劝阻过负载STL功能？

发布于 2025-01-21 07:07:05 字数 900 浏览 1 评论 0原文

我想知道是否不愿超载STL功能，如果是，为什么？

我昨天遇到了一个问题，发现std :: fpClassify在Microsoft Compilers上没有积分超载（ https://learn.microsoft.com/en-us/cpp/cpp/cpp/cpp/c-runtime-library/reference /fpClassify？view = msvc-170 ）与其他编译器一样（请参见 https://en.cppreference.com/w/cpp/numeric/math/math/fpclassify ）。

编译时，我遇到了这个问题

T var; // T can be an integral type
std::isnan(var); //

当我尝试使用Microsoft C ++编译器

。我已经有一个工作解决方案来解决此问题，该问题不涉及过载std :: fpClassify，但是我确实考虑了可能只是为std :: fpclassify我自己写一个过载。，但是看来这本来会变得有毛，因为代码可以使用非Microsoft编译器编译，在这种情况下，我们已经定义了积分超载。

原文

I was wondering if it is discouraged to overload an STL function, and if so, why?

I ran across an issue yesterday, where I found that std::fpclassify doesn't have an integral overload on microsoft compilers (https://learn.microsoft.com/en-us/cpp/c-runtime-library/reference/fpclassify?view=msvc-170) as it does for other compilers (see (4) in https://en.cppreference.com/w/cpp/numeric/math/fpclassify).

I ran across this issue when I tried to compile

T var; // T can be an integral type
std::isnan(var); //

using microsoft C++ compiler.

I already have a working solution to solve this issue that didn't involve overloading std::fpclassify, but I did consider maybe just writing an overload for std::fpclassify myself, but it seems it would have gotten hairy because the code might be compiled using non-microsoft compilers, in which case, we would already have the integral overload defined.

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睫毛溺水了 2025-01-28 07:07:06

否。只要您不试图将自定义超载放在命名空间std中，就没有问题，因为在某些情况下，这是未定义的。

考虑std ::交换甚至鼓励您在交换两个对象时提供超负荷的位置应执行std ::交换无法做的事情：

#include <iostream>
#include <algorithm>

namespace X {

struct foo {
    int value;
    bool operator<(const foo& other) { return value < other.value; }
};

void swap(foo& a, foo& b) noexcept {
    std::cout << "swap\n";
    std::swap(a.value,b.value);
}

}

int main() {
    std::vector<X::foo> v{{2},{1}};
    std::iter_swap(v.begin(),v.begin()+1);   // calls X::swap
    std::cout << v[0].value << " " << v[1].value << "\n";

    std::vector<int> w{2,1};
    std::iter_swap(w.begin(),w.begin()+1);  // calls std::swap
    std::cout << w[0] << " " << w[1];
}

输出：

swap
1 2
1 2

这取决于ADL，这意味着它对std的基本类型或类型都无法使用。仅当您将它们包装在自定义类中的某个名称空间中时，就可以将其用于int等。

No. There is no issue as long as you are not trying to place your custom overload inside the namespace std, because that is undefined except in some cases.

Consider std::swap where you are even encouraged to provide an overload when swapping two objects should do something that std::swap cannot do:

#include <iostream>
#include <algorithm>

namespace X {

struct foo {
    int value;
    bool operator<(const foo& other) { return value < other.value; }
};

void swap(foo& a, foo& b) noexcept {
    std::cout << "swap\n";
    std::swap(a.value,b.value);
}

}

int main() {
    std::vector<X::foo> v{{2},{1}};
    std::iter_swap(v.begin(),v.begin()+1);   // calls X::swap
    std::cout << v[0].value << " " << v[1].value << "\n";

    std::vector<int> w{2,1};
    std::iter_swap(w.begin(),w.begin()+1);  // calls std::swap
    std::cout << w[0] << " " << w[1];
}

Output:

swap
1 2
1 2

This relies on ADL, which means it wont work for fundamental types or types from std. Only if you wrap them inside a custom class inside some namespace like above you can use it for int and the like.

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