[C ++编译时断言]:如果不满足某些条件，我们可以抛出编译错误吗? [英] [C++ compile time assertions]: Can we throw a compilation error if some condition is not met?

问题描述

我写了一个函数:

template<int N> void tryHarder() {
    for(int i = 0; i < N; i++) {
        tryOnce();
    }
}

但是我只希望在N在0到10之间的情况下进行编译.我可以这样做吗?怎么样?

but I only want it to compile if N is in between 0 and 10. Can I do it? How?

推荐答案

您可以使用 static_assert声明进行操作:

You can do it with static_assert declaration:

template<int N> void tryHarder() {

    static_assert(N >= 0 && N <= 10, "N out of bounds!");

    for(int i = 0; i < N; i++) {
        tryOnce();
    }
}

此功能仅在C ++ 11起可用.如果您使用的是C ++ 03，请看看 Boost的静态断言宏.

This feature is only avaliable since C++11. If you're stuck with C++03, take a look at Boost's static assert macro.

这的整个想法都是不错的错误消息.如果您不关心这些东西，或者甚至无法提高自己的能力，则可以执行以下操作:

The whole idea of this are nice error messages. If you don't care for those, or can't even affor boost, you could do something as follows:

template<bool B>
struct assert_impl {
    static const int value = 1;
};

template<>
struct assert_impl<false> {
    static const int value = -1;
};

template<bool B>
struct assert {
    // this will attempt to declare an array of negative
    // size if template parameter evaluates to false
    static char arr[assert_impl<B>::value]; 
};

template<int N>
void tryHarder()
{
    assert< N <= 10 >();
}

int main()
{
    tryHarder<5>();  // fine
    tryHarder<15>();  // error, size of array is negative
}

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