保留顺序时将浮点转换为unsigned int [英] Converting floating point to unsigned int while preserving order

问题描述

我发现了很多关于SO的答案，重点是将 float 转换为 int .

I have found a lot of answers on SO focusing on converting float to int.

我只处理正浮点值.我一直在使用的一种简单方法是:

I am manipulating only positive floating point values. One simple method I have been using is this:

unsigned int float2ui(float arg0) {
    float f = arg0;
    unsigned int r = *(unsigned int*)&f;
    return r;
}

上面的代码运行良好，但是无法保留数字顺序.按顺序，我的意思是这样:

The above code works well yet it fails to preserve the numeric order. By order I mean this:

  float f1 ...;
  float f2 ...;
  assert( ( (f1 >= f2) && (float2ui(f1) >= float2ui(f2)) ) ||
          ( (f1 <  f2) && (float2ui(f1) < vfloat2ui(f2)) ));

我尝试使用具有相同结果的联合.任何想法?我使用Homebrew gcc 5.3.0.

I have tried to use unions with the same results. Any idea? I use Homebrew gcc 5.3.0.

推荐答案

您所编写的代码具有未定义的行为.如果您想半移植地访问 float 的表示形式(实现定义，定义良好并假定IEEE 754，并且float和integer字节序匹配)，则应该执行以下操作:

The code you're using, as writen, has undefind behavior. If you want to access the representation of floats semi-portably (implementation-defined, well-defined assuming IEEE 754 and that float and integer endianness match), you should do:

uint32_t float2ui(float f){
    uint32_t r;
    memcpy(&r, &f, sizeof r);
    return r;
}

对于非负值，浮点值和表示形式之间的映射保留顺序.如果您认为自己无法保留订单，则需要准确查看您认为哪些值是反例.

For non-negative values, this mapping between floating point values and representation is order-preserving. If you think you're seeing it fail to preserve order, we'll need to see exactly what values you think are a counterexample.

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