转换浮动为unsigned long访问浮动内部，在C＃定义

问题描述

我要一个浮动转换为无符号长，同时保持二进制重新presentation在浮动（所以我做的不的希望蒙上 5.0 到 5 ！）。

I want to convert a float to a unsigned long, while keeping the binary representation of the float (so I do not want to cast 5.0 to 5!).

这是很容易在下面的方法做：

This is easy to do in the following way:

float f = 2.0;
unsigned long x = *((unsigned long*)&f)

不过，现在我需要做同样的事情在的#define ，因为我想在某些数组初始化（这样的[在线]功能以后使用是不是一种选择）。

However, now I need to do the same thing in a #define, because I want to use this later on in some array initialization (so an [inline] function is not an option).

这不会编译：

#define f2u(f) *((unsigned long*)&f)

如果我把它称为是这样的：

If I call it like this:

unsigned long x[] = { f2u(1.0), f2u(2.0), f2u(3.0), ... }

我得到的错误是（逻辑）：

The error I get is (logically):

lvalue required as unary ‘&’ operand

---

注意：一个解决方案，建议下面是使用联盟键入我的数组。然而，这是没有选择。我实际上做以下内容：

---

Note: One solution that was suggested below was to use a union type for my array. However, that's no option. I'm actually doing the following:

#define Calc(x) (((x & 0x7F800000) >> 23) - 127)
unsigned long x[] = { Calc(f2u(1.0)), Calc(f2u(2.0)), Calc(f2u(3.0)), ... }

因此​​，阵列真的会/类型必须为长[] 。

推荐答案

您应该使用一个联盟：

union floatpun {
    float f;
    unsigned long l;
};

union floatpun x[3] = { {1.0}, {2.0}, {3.0} };

或者

union {
    float f[3];
    unsigned long l[3];
} x = { { 1.0, 2.0, 3.0 } };

（后者将让你通过 XL ，你需要类型的数组无符号长[3] ）

(The latter will let you pass x.l where you need an array of type unsigned long [3]).

当然，你需要确保无符号长和浮动有你的平台上相同的大小。

Of course you need to ensure that unsigned long and float have the same size on your platform.

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