“整数常数溢出”； constexpr中的警告 [英] "integer constant overflow" warning in constexpr

问题描述

我试图找到一个与constexpr兼容的哈希函数，以用于在编译时对字符串进行哈希处理。字符串的数量确实很少（小于10），而且我需要单独检查冲突，因此该算法可能远非完美。我在互联网上的某个地方找到了以下版本的FNV1A：

I'm trying to find a constexpr compatible hash function to use for hashing strings in compile-time. The number of strings are really small (<10) and I have a separate check for collisions, so the algorithm can be far from perfect. I found the following version of FNV1A somewhere on the internet:

static constexpr unsigned int Fnv1aBasis = 0x811C9DC5;
static constexpr unsigned int Fnv1aPrime = 0x01000193;

constexpr unsigned int hashFnv1a(const char *s, unsigned int h = Fnv1aBasis)
{
    return !*s ? h : hashFnv1a(s + 1, (h ^ *s) * Fnv1aPrime);
}

但是当我在MSVS 2015中进行编译时，会收到以下警告：

But when I compile this in MSVS 2015 I get the following warning:

warning C4307: '*': integral constant overflow

由于函数中只有一个乘法，因此我认为警告来自（h ^ * s）* Fnv1aPrime 。这是有道理的，因为将 0x811C9DC5 （ Fnv1aBasis ）乘以几乎任何东西都会导致32位整数溢出。

Since there's only one multiplication in the function I would assume the warning comes from (h ^ *s) * Fnv1aPrime. It makes sense since multiplying 0x811C9DC5 (Fnv1aBasis) with just about anything will make a 32-bit integer overflow.

有什么办法可以解决此问题？我已经尝试了一些其他的用于对字符串进行哈希处理的constexpr函数，但它们都具有相同的问题。

Is there any way to work around this? I've tried a couple of other constexpr functions I've found for hashing strings, but all of them have the same issue.

推荐答案

您可以明确地强制转换为 unsigned long long 并返回，如下所示：

You can explicitly cast to unsigned long long and back, as follows:

constexpr unsigned int hashFnv1b(const char *s, unsigned int h = Fnv1aBasis)
{
    return !*s
           ? h
           : hashFnv1b(
               s + 1,
               static_cast<unsigned int>(
                 (h ^ *s) * static_cast<unsigned long long>(Fnv1aPrime)));
}

此在我的现场演示中停止警告（第20行触发它，而第21行则不触发）。

This stops the warning in my live demo (line 20 triggers it, line 21 does not).

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