获得必要的最小字节数来保持一个给定的无符号的整数 [英] Get the minimum number of bytes necessary to hold a given unsigned integer

问题描述

我需要写一系列的无符号整数到一个文件，每一个是不超过 N 在运行时确定的限制更大。为了节省空间，我要收拾他们在短短字节可能。但是，我不知道如何计算要追究他们的最小字节数，所以我只有以下，丑陋的解决方案：

I need to write a series of unsigned integers to a file, each one being no greater than a limit n determined at runtime. To save space, I want to pack them in as little bytes as possible. However, I've no idea how to compute the minimum number of bytes necessary to hold them, so I only have the following, ugly solution:

int get_needed_bytes(uint32_t n) {
    if (n < 256) return 1;
    else if (n < 65536) return 2;
    else if (n < 16777216) return 3;
    return 4;
}

有没有更好的方法来达到同样的目的？

Is there a better way to achieve the same purpose?

推荐答案

您可以尝试沿着这些路线的东西（未经测试）。

You might try something along these lines (untested).

int GetNeededBytes(uint32_t n)
{
    // Maximum number of bytes supported
    int bytes = 4;
    // Get mask for highest order byte
    // Warning: watch for overflow here
    // 4 bytes should resolve to 0xff000000
    int mask = 0xff << (bytes * 8);

    while (bytes > 0)
    {
        if (n & mask)
            return bytes;
        mask /= 0x100;
        bytes--;
    }
    return 0;
}

但我不知道为什么这是一个好主意。为了读出值回来，你需要一种方法来标志有多少字节重新present下一个值。我猜想，计数值会带走大部分保存的字节数。

But I'm not sure why this is a good idea. In order to read the values back, you need a way to flag how many bytes represent the next value. I suspect that count value will take away most of the bytes you saved.

有更好的融为一体pression技术可用。

There are better compression techniques available.

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