您可以修饰一个十六进制常量，以便将其解释为浮点数吗？
只是对某件事感到好奇。在将 HTML 颜色（其各个颜色分量通过 2 位十六进制值表示）转换为 0.0 到 1.0 之间的浮点数以便我们可以在 OpenGL 中使用它们时，我对某些事情感到好奇。我们必须将十六进制值除以 255 才能得到相应的 OpenGL 值，但我们不能简单地按原样使用十六进制值，因为这会产生整数除法。
现在我知道这些都可以解决这个问题（当然）...
float r = 0xFD / (float)0xFF; // Works because of the cast to float
float r = 0xFD / 255.0f; // Works because of the explicit float
float d = 0xFF;
float r = 0xFD / d; // Works because 'd' is a float
...但是我想知道是否有任何方法可以装饰十六进制值，以便将其解释为浮点数（就像您对 1.0 中的“f”所做的那样） f) 无需进行转换、计算或临时变量。
这些当然是行不通的...
float r = 0xFD / 0xFF; // Integer division yields an integer, not float
float r = 0xFD / 0xFFf; // Interprets 'f' as part of the hex value
同样，我没有试图找出如何实现我需要的结果，因为我的代码工作得很好。我只是想知道是否可以通过使用类似于“f”处理十进制值的方式装饰十六进制值来使代码更清晰，而不是使用上述三种有效的方法。
Just curious about something. While converting HTML colors (whose individual color components are represented via 2-digit hexadecimal values) to floats between 0.0 and 1.0 so we can use them with OpenGL, I became curious about something. We have to divide the hex values by 255 to give us their OpenGL counterparts, but we can't simply use the hex values as-is because that produces integer division.
Now I know these all work around this issue (of course)...
float r = 0xFD / (float)0xFF; // Works because of the cast to float
float r = 0xFD / 255.0f; // Works because of the explicit float
float d = 0xFF;
float r = 0xFD / d; // Works because 'd' is a float
...but I was wondering if there's any way to just decorate a hex value so it's interpreted as a float (like you do with the 'f' in 1.0f) without having to do casting, calculations or interim variables.
These of course don't work...
float r = 0xFD / 0xFF; // Integer division yields an integer, not float
float r = 0xFD / 0xFFf; // Interprets 'f' as part of the hex value
Again, not trying to find out how to achieve my needed results as my code works just fine. I'm just wondering if I can make the code cleaner via decorating the hex value with something similar to how 'f' works with decimal values instead of using the above-three methods that do work.
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在 c++0x 中，使用 用户定义的文字 可能可以实现这一点但我怀疑在 c++03 中是否有任何巧妙的方法可以做到这一点。
This would likely be possible in c++0x using user defined literals but I doubt there is any neat way to do it in c++03.
恐怕您必须使用预处理器定义才能获得最佳外观。并且您还需要进行转换来执行计算。
我能想到的最好的解决方案是
浮点值不能指定为十六进制（整数）值。
I'm afraid you have to use preprocessor definition in order to achieve the best look. And you need also casting to perform your computations.
The best solution I can think is
Floating point values cannot be specified as hexadecimal (integer) values.
我不知道有什么方法可以完全满足您的要求，但如果您想重复使用 0xFF，为什么不将其设置为常量呢？这仍然会保留打字，即：
我认为这会解决您想要解决的问题，即使它的方式不完全相同。
I don't know of any way to do exactly what you're asking, but if you're looking to use 0xFF repeatedly, why not set it as a constant? That will still preserve typing, i.e:
I think that'd solve the issue you're looking to solve, even if it's not exactly in the same way.
一种方法是定义自己的模式来定义十六进制浮点数，0xff 定义十六进制整数，您可以创建一个宏，例如：
然后您只需编写
即可进行浮点除法。
虽然此解决方案可能会给阅读您代码的其他人造成混乱，但如果代码主要适合您，那么您可以这样做
One way you could do it is by defining your own pattern to define hex floats, 0xff defines hex integers, you could make a macro, like:
Then you just write
to do your float division.
This solution could though create confusion for others who reads your code, but if the code is mostly for you, than you could do it this way